What is the mass, in kilograms, of 3.28 x 103 mol of H2O?

Answers

Answer 1

3.28 x 10³ mol of water has a mass of 59.1156 kg. This is the accurate answer to the given question.

How do you determine 3.28 x 10³ mol of H2O's mass in kilograms?

We can use the molar mass of water, which is 18.02 g/mol, to calculate the mass of 3.28 x 10³ moles of H2O. The following equation can be used to translate the quantity of moles into grams:

mass is determined by multiplying the number of moles by the molar mass.

Inputting the values provided yields:

18.02 g/mol times 3.28 x 10 3 moles of mass equals 59,115.6 g.

Lastly, by dividing by 1000, we can change the mass from grams to kilograms:

mass is 59,115.6 grams per kilogram (59.1156 kg).

As a result, 3.28 x 10³ moles of H2O have a mass of 59.1156 kg.

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Related Questions

_________________________________ are scientists who study the processes that make cells work.

Answers

Answer:

Microbiologists

Explanation:

Microbiologists are scientists who study the processes that make cells work.

I am mixing vinegar and baking soda in a bottle. If I want to be sure that Conservation of Mass is followed, what should I make sure to do?

Answers

Law of conservation of mass was crucial to progression of chemistry, as it helped scientists understand that substances did not disappear as result of reaction.

What is conservation of mass?

Law of conservation of mass explains that in chemical reaction mass is neither created nor destroyed.

To ensure Conservation of Mass is followed when mixing vinegar and baking soda in a bottle, you should make sure to do the following:

Weigh the bottle and its contents before adding vinegar and baking soda. This will give the initial mass of system.

Carefully measure the amounts of vinegar and baking soda you add to bottle.

Quickly and securely cap the bottle after adding baking soda to avoid any gas from escaping.

Weigh the bottle and its contents again after reaction has completed and the gas has stopped escaping. This will give the final mass of the system.

Compare the initial and final masses of system. They should be equal within experimental error.

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To earn full credit for your answers, you must show the appropriate formula, the correct substitutions , and your answer including the correct units.

In 2013, there were 2.6 x 106 people living in Paris. If 7.34 x 104 babies were born what was the crude birth rate?

Answers

The birth rate is 28.2 births per year

What is the birth rate?

The number of births per 1,000 persons in a particular population over a given time period is commonly used to determine the birth rate.

The following is the formula that can be used to calculate the birth rate:

(Number of births / Population) x 1,000 is the birth rate.

The birth rate can be obtained from;

Birth rate = (Number of births / Population) x 1,000

We know that the birth rate can be obtained from;

7.34 x 10^4 / 2.6 x 10^6 * 1000

= 28.2 births per year

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Predict whether each of the following reactions will occur in aqueous solutions. If you predict a reaction will occur, state the products formed and their states (s, l, or aq) and write a balanced chemical reaction. If you predict that a reaction will not occur, state your reasoning. Note: barium sulfate and silver bromide are insoluble salts.

a. Sodium hydroxide(aq) + ammonium sulfate(aq)
b. Niobium(V) sulfate(aq) + barium nitrate(aq)

Answers

a. Sodium hydroxide(aq) + ammonium sulfate(aq) will react to form solid ammonium hydroxide and sodium sulfate(aq). b. Niobium(V) sulfate(aq) + barium nitrate(aq) will react to form solid barium sulfate(s) and niobium(V) nitrate(aq).

What is the significance of predicting whether a reaction will occur in aqueous solution?

Predicting whether a reaction will occur in aqueous solution is important in understanding chemical reactions and their potential outcomes. It allows chemists to determine what products will form and whether a reaction is feasible or not based on the solubility of the reactants and products.

How do you predict whether a reaction will occur in aqueous solution?

To predict whether a reaction will occur in aqueous solution, one can use solubility rules to determine whether the products of the reaction will form an insoluble precipitate or remain in solution. If a precipitate forms, it indicates that a reaction has occurred, whereas if all the products remain in solution, the reaction will not occur. Additionally, one can use other chemical principles, such as redox reactions and acid-base reactions, to predict whether a reaction will occur.

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What is one possible reason dolphin dissapered

Answers

Explanation:

Foremost among the threats is fishing bycatch—marine animals incidentally caught and killed in fishing operations.

This is a serious crisis, not only for these charismatic animals, but also for the health of our oceans, where cetaceans play an irreplaceable role as apex predators.

6) What is the vapor pressure of water at 100°C?

Answers

Answer: The vapor pressure of pure water at 100°C is 760mm of Hg.

Explanation: We know tat 100 degree Celsius the water will convert into gas phase. At boiling point the vapor pressure of any liquid is 1 atmosphere and if we boil the water at 100° Celsius it means 1 atmosphere . So, 1 atmosphere = 760mm of Hg.

How many grams of NH3
form when 22.3 L
of H2(g)
(measured at STP) reacts with N2
to form NH3
according to this reaction?

N2(g)+3H2(g)→2NH3(g)

Answers

Answer:

11.2823 grams of NH3 are produced
Explanation:

No. of moles for H2 = 22.3/22.4 = 0.9955 moles

By calculating number of moles produced of NH3, by using ratios:

N2 + 3H2 → 2NH3
1 : 3 : 2
? : 0.9955: ?

part value = 0.9955/3 = 0.33183
No. of moles for NH3 = 2 * 0.33183 = 0.6637 moles

mass of produced NH3 (Molar mass = 17 g/mol) = 0.6637*17 = 11.2823 grams

an impure sample of table salt that weighed 0.8421 g when dissolved in water and treated with excess agno3 formed 2.044 g of agcl. what percentage of nacl is in the impure sample?

Answers

The percentage of NaCl in the impure sample is 76.25%.

The given weight of an impure sample of table salt is 0.8421 g, and the weight of AgCl formed is 2.044 g. The percentage of NaCl in the impure sample can be determined as follows:

Calculations: Gram molecular weight of NaCl = 23 + 35.5 = 58.5 g

Number of moles of AgCl = weight of AgCl / gram molecular weight of AgCl= 2.044 / 143.5 = 0.0142 mol

The equation of the reaction between NaCl and AgNO3 is given as follows:

NaCl + AgNO3 → NaNO3 + AgCl

It can be seen from the above equation that 1 mole of NaCl gives 1 mole of AgCl.So, the number of moles of NaCl present in the sample is 0.0142 mol.

Percentage of NaCl in the sample = number of moles of NaCl present in the sample / number of moles of the impure sample × molar mass of NaCl × 100= (0.0142 / (0.8421 / 58.5)) × 100= 76.25%

Therefore, the percentage of NaCl in the impure sample is 76.25%.

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PLEASE HELP
It's for Modeling Electron Configurations

Answers

The valence electrons of oxygen are; 2s2 2p4

The valence electrons of nitrogen are 2s2 2p3

This is how we can be able to know the number of valence electrons.

What are the valence electrons of Nitrogen and oxygen?

Valence electrons are the electrons in the outermost shell of an atom. They are the electrons involved in chemical reactions and bonding with other atoms. In the case of nitrogen, its electron configuration is 1s^2 2s^2 2p^3, which means it has two electrons in the first shell, two electrons in the second shell, and three electrons in the outermost p orbital, giving it a total of 5 valence electrons.

Oxygen has an electron configuration of 1s^2 2s^2 2p^4, which means it has two electrons in the first shell, two electrons in the second shell, and four electrons in the outermost p orbital, giving it a total of 6 valence electrons.

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Please help and write full solution

Answers

The pH of the solution is 1.8. Option C is the right answer

How The answer was obtained

First, we need to calculate the moles of LiOH and HNO3:

Moles of LiOH = 0.1 mol/L × 0.025 L = 0.0025 mol

Moles of HNO3 = 0.1 mol/L × 0.035 L = 0.0035 mol

Next, we need to determine which of the reactants will be limiting. LiOH and HNO3 react in a 1:1 stoichiometric ratio, so whichever reactant is present in the smaller amount will be limiting. In this case, LiOH is limiting because it has fewer moles than HNO3.

The balanced chemical equation for the reaction between LiOH and HNO3 is:

LiOH + HNO3 → LiNO3 + H2O

Since LiOH is limiting, all of it will react with HNO3. This means that all of the moles of HNO3 will be consumed, and there will be some excess LiOH remaining.

Moles of LiOH remaining = Moles of LiOH - Moles of HNO3 = 0.0025 mol - 0.0035 mol = -0.0010 mol

Note that the result is negative, which means that all of the HNO3 has been consumed, and there is still some LiOH remaining.

The concentration of LiOH in the final solution is:

[LiOH] = Moles of LiOH remaining / Total volume of solution = -0.0010 mol / (0.025 L + 0.035 L) = -0.010 mol/L

Again, the result is negative, which is not physically meaningful. This indicates that our assumption that LiOH is limiting is incorrect.

To correct this, we can assume that HNO3 is limiting and repeat the calculations. In this case, all of the HNO3 will react with LiOH, and there will be some excess HNO3 remaining.

Moles of HNO3 remaining = Moles of HNO3 - Moles of LiOH = 0.0035 mol - 0.0025 mol = 0.0010 mol

The concentration of HNO3 in the final solution is:

[HNO3] = Moles of HNO3 remaining / Total volume of solution = 0.0010 mol / (0.025 L + 0.035 L) = 0.01 mol/L

The balanced chemical equation for the reaction between HNO3 and LiOH is:

HNO3 + LiOH → LiNO3 + H2O

The stoichiometry of this reaction tells us that one mole of HNO3 reacts with one mole of LiOH to produce one mole of water. Therefore, the moles of water produced in the reaction are equal to the moles of HNO3 remaining:

Moles of water = 0.0010 mol

The volume of the final solution is:

Volume = 0.025 L + 0.035 L = 0.06 L

The molarity of the water in the final solution is:

Molarity of water = Moles of water / Volume of solution = 0.0010 mol / 0.06 L = 0.0167 mol/L

The pH of water is 7.0, and the pH of a solution is related to the concentration of H+ ions in the solution. Since water is neutral and does not contribute any H+ ions, the pH of the final solution is:

pH = 7.0 - log[H+] = 7.0 - log(0.0167) ≈ 1.8

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Determine the pH of the solution when 25ml of 0.1mLiOH is added to 35ml of 0.1m nitric acid

(A) 0.23

(B) 0.47

(D) 7.0

Two elements, and , combine to form two binary compounds. In the first compound, 23. 3 g of combines with 3. 00 g of. In the second compound, 7. 00 g of combines with 4. 50 g of. Show that these data are in accord with the law of multiple proportions. If the formula of the second compound is , what is the formula of the first compound?

Compound I:

Answers

The formula for the first Compound I is likely A2B, or O2N.

To show that the data are in accordance with the law of multiple proportions, we need to calculate the ratios of the masses of element B that combine with a fixed mass of element A in each compound. If these ratios are in simple whole-number ratios, then the law of multiple proportions is upheld.

For Compound I:

Mass of A (23.3 g) / Mass of B (3.00 g) = 7.77

For Compound II:

Mass of A (7.00 g) / Mass of B (4.50 g) = 1.56

These ratios are not equal, indicating that the two compounds have different ratios of A and B. However, we can simplify the ratio for Compound II by dividing both masses by the mass of B:

Mass of A (7.00 g / 4.50 g) = 1.56

This suggests that Compound II has a 2:3 ratio of A to B.

Using this information, we can determine the formula of Compound II. Let x be the atomic mass of A and y be the atomic mass of B:

(2x) + (3y) = 4.50 g

Solving for y gives us y = 1.00 g/mol. Since the formula of Compound II is AB2, we know that the atomic mass of A is 2 g/mol. Therefore, A is likely oxygen (atomic mass 16 g/mol).

For Compound I, we know that A is oxygen (atomic mass 16 g/mol). Using the same approach as above, we can determine that B has an atomic mass of 14 g/mol. Therefore, the formula for Compound I is likely A2B, or O2N.

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Name the following formula: BF2 (hint: treat boron like a nonmetal)
A. Boron fluoride
B. Monoboron difluoride
C. Boron difluoride
D. Boron (II) fluoride
E. Best Friends Forever

Answers

The name of the formula BF2 is : C.) Boron difluoride. The number of atoms of each element in a compound is known by chemical formula.

What is meant by chemical formula?

Chemical formula is an expression that shows elements in a compound and relative proportions of those elements. If only one atom of specific type is present, then no subscript is used. For atoms that have two or more of a specific type of atom present, then a subscript is written after the symbol for that atom.

Chemical formula contains the symbols of atoms of the elements present in the compound and also how many there are for each element in the form of subscripts.

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Calculate the pH at the following points in a titration of 40 mL (0. 040 L) of 0. 100 M

barbituric acid (Ka = 9. 8  10–5) with 0. 100 M KOH.

(a) no KOH added

(b) 20 mL of KOH solution added

(c) 39 mL of KOH solution added

(d) 40 mL of KOH solution added

(e) 41 mL of KOH solution added

Answers

We are given:

Molarity of
C
4
H
4
N
2
O
3
=
0.100

M
Volume of
C
4
H
4
N
2
O
3
=
40

m
L
Molarity of
K
O
H
=
0.100

M
K
a
=
9.8
×
10
−
5

Chemical reaction:
C
4
H
4
N
2
O
3
+
K
O
H
→
C
4
H
3
N
2
O
3
K
+
H
2
O

Part a:
no KOH is added.

The concentration of hydrogen ion can be calculated as:

C
4
H
4
N
2
O
3
⇌
C
4
H
3
N
2
O
−
3
+
H
+
K
a
=
[
C
4
H
3
N
2
O
−
3
]
[
H
+
]
[
C
4
H
4
N
2
O
3
]
[
C
4
H
3
N
2
O
−
3
]
=
[
H
+
]
9.8
×
10
−
5
=
[
H
+
]
2
[
C
4
H
4
N
2
O
3
]
9.8
×
10
−
5
=
[
H
+
]
2
0.100
[
H
+
]
2
=
9.8
×
10
−
6
[
H
+
]
=
3.13
×
10
−
3

M

The pH value of the solution is:

p
H
=
−
log
[
H
+
]
p
H
=
−
log
(
3.13
×
10
−
3
)
p
H
=
2.50

Part b:
20 mL of KOH is added.

We will find the initial moles of the reactants:

M
o
l
e
s
(
n
)
=
M
o
l
a
r
i
t
y
(
M
)
×
V
o
l
u
m
e
(
V
)
n
C
4
H
4
N
2
O
3
=
0.100

M
×
0.040

L
=
0.0040

m
o
l
n
K
O
H
=
0.100

M
×
0.020

L
=
0.0020

m
o
l

Moles of the acid and salt after the addition of KOH:

n
C
4
H
4
N
2
O
3
=
0.00400
−
0.0020
=
0.0020

m
o
l
n
C
4
H
3
N
2
O
3
K
=
0.00200

m
o
l

The concentration of acid and salt:

M
=
n
V
V
=
40
+
20
=
60

m
L
[
C
4
H
4
N
2
O
3
]
=
0.0020

m
o
l
0.060

L
=
0.033

M
[
C
4
H
3
N
2
O
3
K
]
=
0.0020

m
o
l
0.060

L
=
0.033

M

The pH value is calculated as:

p
H
=
−
log
(
K
a
)
+
log
(
[
C
4
H
3
N
2
O
3
K
]
[
C
4
H
4
N
2
O
3
]
)
p
H
=
−
log
(
9.8
×
10
−
5
)
+
log
(
0.033
0.033
)
p
H
=
4.01

Part c:
39 mL of KOH is added.

We will find the initial moles of the reactants:

M
o
l
e
s
(
n
)
=
M
o
l
a
r
i
t
y
(
M
)
×
V
o
l
u
m
e
(
V
)
n
C
4
H
4
N
2
O
3
=
0.100

M
×
0.040

L
=
0.0040

m
o
l
n
K
O
H
=
0.100

M
×
0.039

L
=
0.0039

m
o
l

Moles of the acid and salt after the addition of KOH:

n
C
4
H
4
N
2
O
3
=
0.00400
−
0.0039
=
0.0001

m
o
l
n
C
4
H
3
N
2
O
3
K
=
0.0039

m
o
l

The concentration of acid and salt:

M
=
n
V
V
=
40
+
39
=
79

m
L
[
C
4
H
4
N
2
O
3
]
=
0.0001

m
o
l
0.079

L
=
0.00126

M
[
C
4
H
3
N
2
O
3
K
]
=
0.0039

m
o
l
0.079

L
=
0.0494

M

The pH value is calculated as:

p
H
=
−
log
(
K
a
)
+
log
(
[
C
4
H
3
N
2
O
3
K
]
[
C
4
H
4
N
2
O
3
]
)
p
H
=
−
log
(
9.8
×
10
−
5
)
+
log
(
0.0494
0.00126
)
p
H
=
4.01
+
1.59
p
H
=
5.60

Part d:
40 mL of KOH is added.

We will find the initial moles of the reactants:

M
o
l
e
s
(
n
)
=
M
o
l
a
r
i
t
y
(
M
)
×
V
o
l
u
m
e
(
V
)
n
C
4
H
4
N
2
O
3
=
0.100

M
×
0.040

L
=
0.0040

m
o
l
n
K
O
H
=
0.100

M
×
0.040

L
=
0.0040

m
o
l

Moles of the acid and salt after the addition of KOH:

n
C
4
H
4
N
2
O
3
=
0.00400
−
0.0040
=
0.0

m
o
l
n
C
4
H
3
N
2
O
3
K
=
0.0040

m
o
l

The concentration of acid and salt:

M
=
n
V
V
=
40
+
40
=
80

m
L
[
C
4
H
4
N
2
O
3
]
=
0.0

m
o
l
0.080

L
=
0.0

M
[
C
4
H
3
N
2
O
3
K
]
=
0.0040

m
o
l
0.080

L
=
0.050

M

The pH value is calculated as:

p
H
=
7
+
1
2
(
p
K
a
+
log
[
C
4
H
3
N
2
O
3
K
]
)
p
H
=
7
+
1
2
(
4.01
+
log
(
0.05
)
)
p
H
=
7
+
1
2
(
4.01
−
1.30
)
p
H
=
7
+
1.35
p
H
=
8.35

Part e:
41 mL of KOH is added.

We will find the initial moles of the reactants:

M
o
l
e
s
(
n
)
=
M
o
l
a
r
i
t
y
(
M
)
×
V
o
l
u
m
e
(
V
)
n
C
4
H
4
N
2
O
3
=
0.100

M
×
0.040

L
=
0.0040

m
o
l
n
K
O
H
=
0.100

M
×
0.041

L
=
0.0041

m
o
l

Moles of the base remains left over after the addition of KOH:

n
K
O
H
=
0.0041
−
0.0040
=
0.0001

m
o
l

The concentration of base:

M
=
n
V
V
=
40
+
41
=
81

m
L
[
K
O
H
]
=
0.0001

m
o
l
0.081

L
=
0.00123

M

The concentration of hydroxide ion is equal to 0.00123 M because it dissolves as:

K
O
H
→
K
+
+
O
H
−

The pOH value of the solution:

p
O
H
=
−
log
[
O
H
−
]
p
O
H
=
−
log
(
0.00123
)
p
O
H
=
2.91

The pH value is calculated as:

p
H
+
p
O
H
=
14
p
H
+
2.91
=
14
p
H
=
11.1

What is the percent composition of MgSO4 in a solution if 18.6 grams MgSO4 is dissolved in 229 g water ?

Answers

Percent of water of crystallisation in MgSO4⋅7H2O=2467×18×100=51. 2 %.

How can you tell what proportion of the composition is water?

Divide by 100 the mass of the hydrate and the volume of water lost. The amount of water in one mole of the hydrate must be multiplied by 100 after being divided by the hydrate's molar mass in order to determine the theoretical (actual) percent hydration (percent water) using the formula.

Hence, this ratio might be represented by one magnesium sulfate formula unit to seven water molecules. Since MgSO47H2O is the chemical formula for magnesium sulfate hydrate, the value of x must be seven. The value of x in the chemical formula for magnesium sulfate hydrate is thus seven.

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is the diamond cubic structure close-packed? explain your rationale. note: a close-packed structure is one where you can find a plane of atoms in that structure where all the atoms touch. for example, fcc and hcp are close- packed structures.

Answers

Yes, the diamond cubic structure is close-packed. In a close-packed structure, you can find a plane of atoms where all the atoms touch. To understand why the diamond cubic structure is close-packed, follow these steps:

1. Identify the structure: The diamond cubic structure consists of two interpenetrating face-centered cubics (fcc) lattices with a relative shift of 1/4 along the space diagonal.

2. Locate the planes: In the diamond cubic structure, there are several planes where atoms touch each other, forming a hexagonal close-packed (hcp) arrangement. For instance, the (111) plane in one of the fcc sublattices is close-packed.

3. Examine neighboring atoms: In these close-packed planes, the atoms are arranged so that each atom touches its neighboring atoms. This close arrangement of atoms maximizes the packing efficiency.

So, based on the presence of close-packed planes and the neighboring atoms touching each other, the diamond cubic structure is considered a close-packed structure.

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At the carnival, a baby was given a helium balloon whose pressure equaled the

atmospheric pressure at the carnival which was 102. 7 kPa. However, the baby wasn't

paying attention and the balloon floated away into the troposphere where the

atmospheric pressure is 97. 8 kPa. What is the positive difference between the

pressure in the balloon and the pressure of the troposphere in atm?

atm

Answers

The positive difference between the

pressure in the balloon and the pressure of the troposphere is 0.048 kPa .

The balloon is being compressed by atmospheric force. The balloon is trying to inflate owing to gas pressure. The atmospheric pressure (outside pressure) in a fully inflated balloon is lower than the internal gas pressure.

An inflated balloon has a higher internal pressure than its surrounding air. Because the contracting force produced by the rubber's surface tension balances the expanding force caused by the pressure differential, the balloon remains stable.

Because to the air pressure on the walls, a balloon's surface feels tighter when it is inflated. The internal air pressure gives the balloon its shape. The balloon will therefore be more pressurized after being inflated.

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(d) If 48.0 g of NaCl react with 19.0 g of H₂SO4, what mass of Na2SO4 will be produced?

Answers

Answer:

the mass of Na2SO4 produced is 27.6 g.

Explanation:

determine the balanced chemical equation for the reaction between NaCl and H₂SO4, and then use stoichiometry to calculate the mass of Na2SO4 produced.

The balanced chemical equation for the reaction is:

2 NaCl + H₂SO4 → Na2SO4 + 2 HCl

This equation tells us that 2 moles of NaCl react with 1 mole of H₂SO4 to produce 1 mole of Na2SO4.

First, we need to determine the number of moles of NaCl and H₂SO4 we have:

moles of NaCl = mass / molar mass = 48.0 g / 58.44 g/mol = 0.821 mol

moles of H₂SO4 = mass / molar mass = 19.0 g / 98.08 g/mol = 0.194 mol

Since H₂SO4 is the limiting reactant (it is present in less amount than required for complete reaction), we need to use the number of moles of H₂SO4 to calculate the number of moles of Na2SO4 produced:

moles of Na2SO4 = moles of H₂SO4 / 1 x (1 mole Na2SO4 / 1 mole H₂SO4) = 0.194 mol x (1/1) = 0.194 mol

Finally, we can use the molar mass of Na2SO4 to calculate the mass of Na2SO4 produced:

mass of Na2SO4 = moles of Na2SO4 x molar mass = 0.194 mol x 142.04 g/mol = 27.6 g

Therefore, the mass of Na2SO4 produced is 27.6 g.

Is the pattern of atomic radius absolute or general (always true or generally true)?

Answers

Answer:

Generally true.

Explanation:

As you go down the columns of the Periodic Table, the radius of the atom increases.

As you go across the Periodic Table (left to right), the radius of the atoms decrease.

Therefore, the second element on the table, Helium, has the smallest radius and element 87, Francium has the largest radius.

There are a few exception scattered throughout the table, such as the lanthanides and actinides, but generally speaking the radius increases as you travel from right to left and top to bottom.

How many moles are there in 5.6x1015 particles of hydrogen peroxide?

Answers

Answer:

0.9299 x [tex]10^-7[/tex] moles of [tex]OH^-[/tex]

Explanation:

[tex]OH^-[/tex]: 5.6 x 10^(15) (brainly doesn't allow exponents greater than 9)

This number of particles can also be represented as 5.6e15.

'e' in this case means 'times 10 to the n' where n is the number that follows e.

Use stoichiometry to convert between different units.

[tex]\frac{5.6e15 particles}{1 sample} *\frac{1 mole}{6.022 e22 particles}= 0.9299 e -7 moles[/tex]

So in that sample of particles, there are 0.9299 x [tex]10^-7[/tex] moles of [tex]OH^-[/tex].

if a given amount of gas has a pressure of 787.0mmHg and a volume of 970.0mL what would its final volume be at 897.0mmHg and keeping the temperature constant

Answers

Final volume of the gas would be 850.5 mL when the pressure is increased from 787.0 mmHg to 897.0 mmHg while keeping the temperature constant.

What is Boyle's law?

Boyle's Law explains that when temperature of given mass of confined gas is constant, then the product of its pressure and volume is also constant.

Given, P1 = 787.0 mmHg ; V1 = 970.0 mL; P2 = 897.0 mmHg

V2 = ?

As P1V1 = P2V2

V2 = P1V1/ P2

V2 = (787.0 mmHg x 970.0 mL) / 897.0 mmHg

V2 = 850.5 mL

Therefore, the final volume of the gas would be 850.5 mL when the pressure is increased from 787.0 mmHg to 897.0 mmHg while keeping the temperature constant.

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Show how to carry out the following transformation in the highest yield possible. Select the appropriate reagents and/or draw the correct organic product at each step. Use an expanded octet on sulfur to minimize formal charges

Answers

Factors that make the higher yield of crop possible are as follows: enough water for irrigation. access of organic manure of high quality . Fertilizers are a factor in the higher yields of high-cost farming.

Fertilizer is used to ensure good vegetative growth (leaves, branches and flowers), giving rise to healthy plants. They are artificially made but have a rapid response. Overuse of fertilisers can cause negative effects like loose soil fertility. There's a risk with any bond that the issuing company might not be able to meet its obligations.

However, the risks of default are typically higher for companies that issue high-yield bonds. Interest rate . If the octet is expanded with preferred number of bonds, formal charge = 0. The S-atom will gain two electrons in order to complete its octet.

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A sample of gas of mass 2.929g occupies a volume of 426mL at 0°C and 1.00atm pressure.what is molecular weight of the gas?

Answers

Answer:

154 g/mole

Explanation:

We are given the mass of the gas, but we also need the number of moles the 2.929g represents. Since we are provided the conditions of the gas, we can the Ideal Gas law to find the number of moles of the mystery gas.

Ideal Gas Law: PV = nRT, where P, V, and T are the pressure, volume, and temperature (temperature must be in degrees Kelvin), n is the moles, and R is the gas constant.

Let's choose the gas constant that has the same units as we were given. R = 0.0820575 [L⋅atm⋅/(K⋅mol)] comes the closest, but we'll still need to convert ml to liters(L) and °C to °K:

426mL = 0.426L

0°C = 273.25 [add 273.15 to the Centrigrade value]

Let's rearrnage the ideal gas law to solve for n, the number of moles:

n = (PV/RT)

Now enter the data:

n = (1atm)(0.426L)/[(0.0820575 L⋅atm⋅/(K⋅mol))*(273.15°K)]

n = (1atm)(0.426L)/[(0.0820575 L⋅atm⋅/(K⋅mol))*(273.15°K)] [Units that cancle are highlighted]

n = (1)(0.426)/[(0.0820575 /(mol))*(273.15)] [We are left only with moles (mol)

n = (0.426)/(0.0820575)/(273.15) [1/1/mol] [Move the only unit out (1/1/mol)]

n = (0.426)/(0.0820575)/(273.15) [1/1/mol] = 0.0190 moles

Note that the unit moves to the top, i.e., : 1/1/mol = mole

We have the mass and the number of moles. Divide the two to obtain molar mass:

(2.929g)/(0.0190 moles) = 154 g/mole This is also the molecular weight.

[I don't know what is a gas at 0°C and has that molecular weight]

An aerosol deodorant can has a pressure of 3.00 atm at 25 °C. What is the pressure inside the can at a temperature of 845 °C? This example illustrates why you shouldn’t incinerate aerosol cans.

Answers

The pressure inside the aerosol can at 845°C is 33.8 atm. This is why it is extremely dangerous to incinerate aerosol cans, as they can explode when exposed to high temperatures.

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas:

(P1 x V1) / T1 = (P2 x V2) / T2

where P1, V1, and T1 are the initial pressure, volume, and temperature of the gas, and P2, V2, and T2 are the final pressure, volume, and temperature.

We are given the following:

Initial pressure P1 = 3.00 atm

Initial temperature T1 = 25°C = 298 K

We need to find the final pressure P2 at T2 = 845°C = 1118 K.

Since the can is sealed, volumes V1 and V2 remain the same.

Plugging these values into the combined gas law, we get:

(P1 x V1) / T1 = (P2 x V2) / T2

(3.00 atm x V1) / 298 K = (P2 x V1) / 1118 K

Simplifying this equation, we get the following:

P2 = (3.00 atm x V1 x 1118 K) / (298 K x V1)

P2 = 33.8 atm

Therefore, the pressure inside the aerosol can at 845°C is 33.8 atm, which is significantly higher than at room temperature. This demonstrates why it is extremely dangerous to incinerate aerosol cans, as they can explode when exposed to high temperatures.

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If a 100.0g sample of water at 27.8 degrees Celsius is added to a 100.0g sample of water at 73.7 degrees Celsius determine the final temperature of the water assume no heat is lost to the surroundings

Answers

The final temperature of the water is 32.3°C.

If 500.0 g of water at 80.0°C is mixed with 300.0 g of water at 20.0°C, what is the final temperature of the mixture assuming no heat is lost to the surroundings?

The final temperature of the mixture is 53.3°C.

How much heat is lost by a 50.0 g piece of copper initially at 200.0°C when it is placed in 100.0 g of water initially at 25.0°C? The final temperature of the copper and water is 50.0°C.

The heat lost by the copper is 4.19 kJ.

To determine the final temperature of the water, we can use the principle of conservation of energy. The total heat gained by the cooler water is equal to the total heat lost by the hotter water. The heat gained or lost by a substance can be calculated using the formula:

Q = mcΔT

where Q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

Let's assume that the final temperature of the water is T. Then, the heat lost by the hotter water is:

Q1 = mcΔT = (100.0 g) x (4.184 J/g°C) x (73.7°C - T)

The heat gained by the cooler water is:

Q2 = mcΔT = (100.0 g) x (4.184 J/g°C) x (T - 27.8°C)

Since the total heat gained is equal to the total heat lost, we can equate Q1 and Q2:

(100.0 g) x (4.184 J/g°C) x (73.7°C - T) = (100.0 g) x (4.184 J/g°C) x (T - 27.8°C)

Simplifying the equation:

3134.4 - 41.84T = 418.4T - 11702.72

459.24T = 14837.12

T = 32.3°C

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11. Which one of the following contains the same number of atoms as 7 g

of iron?

(a) 4 g of aluminium (b) 4g of magnesium (C) 4g of sulphur

(d) 3 g of carbon (e) 4 g of calcium

Answers

Answer:

4 grams of sulfur contains the same number of atoms as 7 g of iron

Explanation:

The statement "same number of atoms" is the same as stating "the same numer of moles." Moles are a count of atoms/molecules/etc.

So let's convert the masses into moles. Take the mass of an element and divide that by its molar mass (g/mole) to obtain moles of that element.

See the attached worksheet for these calculations.

The 7 grams of iron represents 0.125 moles of iron. The worksheet tells us that 4 grams of sulfur is also 0.125 moles. Nothing else has that number.

7 g of iron contains the same number of atoms as 4 grams of sulfur.

[1 mole = 6.02x10^23 particles].

Find the volume in liters of a 1. 80 m solution of potassium chloride is required to prepare

0. 100 l of a 0. 600 m solution

Answers

the volume of a 1.80 M solution of potassium chloride that is required to prepare 0.100 L of a 0.600 M solution is 0.0333 L, or 33.3 mL.

To determine the volume of a 1.80 M solution of potassium chloride that is required to prepare 0.100 L of a 0.600 M solution, we can use the formula:

M1V1 = M2V2

where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

Let's use this formula to solve the problem:

M1 = 1.80 M (since we are starting with a 1.80 M solution)

V1 = ? (what we are trying to find)

M2 = 0.600 M (since we want to prepare a 0.600 M solution)

V2 = 0.100 L (since we want to prepare 0.100 L of the final solution)

Substituting these values into the formula, we get:

1.80 M x V1 = 0.600 M x 0.100 L

Simplifying, we get:

V1 = (0.600 M x 0.100 L) / 1.80 M

V1 = 0.0333 L, or 33.3 mL

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The particles in.......... can be separated from
heterogeneous mixtures by passing the mixture through a filter.
-suspension
-solution
-colloid
-pure substance

Answers

The particles in suspension can be separated from heterogenous mixtures by passing the mixture through a filter. Option 1.

Particles in a suspension

The particles in suspension can be separated from heterogeneous mixtures by passing the mixture through a filter.

A suspension is a heterogeneous mixture in which the particles are large enough to settle out over time and can be separated by physical means such as filtration.

Other options such as solution, colloid, and pure substances cannot be separated using a filter.

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what does the NaBr + CI2 in the reaction?
NaBr + CI2 -> NaCI + Br2
A. Reactants
B. States of matter
C. Products
D. Yields

Answers

Answer:- NaBr and Cl2 are reactants .

The atoms/molecules/ions on RHS of the reaction which react to give products are called reactants.

So here Sodium bromide and chlorine are the reactants.

Option A is the correct choice.

Additional information:-

Law of conservation of Mass :-

According to this law , the mass of reactants is equal to the mass of the products . Take a chemical reaction ,

C + O2 ==> CO2

Now suppose that if 12g of carbon reacts with 16g of oxygen to form carbon dioxide, then what will be the mass of carbon dioxide? So by law of conservation of Mass we can say that mass of carbon dioxide will be equal to mass of the reactants that will be 12g + 16g = 28g

With this I am leaving a sample problem for you to solve,

Question: Consider the following reaction .

2C + O2 ==> 2CO

If the mass of carbon taken is 24g and the amount of carbon monoxide formed is 56g , find the amount of oxygen used in the reaction.

Answer:D. Yields

Explanation:

Cite the conditions to make the bulb light up

Answers

In order to make a bulb light up, certain conditions need to be met. Firstly, there must be a closed circuit, meaning that there is a continuous flow of current through the circuit. This can be achieved by connecting the bulb to a power source, such as a battery or a generator, and completing the circuit with wires.

Secondly, the bulb must be connected properly to the circuit. This involves connecting the positive and negative terminals of the bulb to the corresponding terminals of the power source or the wires.

Finally, the bulb must have the correct voltage and wattage rating. Using a bulb with a higher or lower rating than the power source can cause the bulb to either not light up or burn out quickly.

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Zinc hydroxide, Zn(OH)2, is precipitated from the tailings ponds of zinc mining operations before the water is released into local rivers. This reduces the concentration of Zn2+ ions in the wastewater. 5.00 kg of sludge containing zinc hydroxide is titrated against HCl solution and 1.50 L of 2.0 M acid is required to reach the endpoint and neutralize the zinc hydroxide. What mass of zinc is contained in the 5.00 kg sludge?

Answers

Answer:

First, we need to find the number of moles of HCl used in the titration:

moles HCl = M x V = 2.0 mol/L x 1.50 L = 3.00 mol

Since zinc hydroxide and HCl react in a 1:2 molar ratio according to the balanced chemical equation:

Zn(OH)2 + 2 HCl → ZnCl2 + 2 H2O

we know that twice as many moles of HCl are required to react with each mole of Zn(OH)2. Therefore, the number of moles of zinc hydroxide in the 5.00 kg sludge can be calculated as follows:

moles Zn(OH)2 = 1/2 x moles HCl = 1/2 x 3.00 mol = 1.50 mol

Finally, we can use the molar mass of Zn(OH)2 to convert the number of moles to mass:

mass Zn(OH)2 = moles Zn(OH)2 x molar mass Zn(OH)2

= 1.50 mol x 99.39 g/mol

= 149.1 g

Therefore, the mass of zinc contained in the 5.00 kg sludge is 149.1 grams.

Explanation:

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