a sample of oxygen occupied 568 ml. when the pressure increased to 911.6 mm Hg. at constant temperature, what volume did the gas initially occupy when the pressure was 248.71 mm Hg?
a 2082
b 155.0
c 2114
d 399.2

Answers

Answer 1

Answer:

[tex]\boxed {\boxed {\sf A. \ 2082 \ mL}}[/tex]

Explanation:

We are asked to find the initial volume of a gas given a change in pressure. Since the temperature remains constant, we are only concerned with volume and pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula for this law is:

[tex]P_1V_1= P_2V_2[/tex]

The pressure was initially 248.71 millimeters of mercury, but the volume is unknown.

[tex]248.71 \ mm \ Hg * V_1 = P_2V_2[/tex]

The pressure is increased to 911.6 millimeters of mercury and the volume is 568 milliliters.

[tex]248.71 \ mm \ Hg * V_1 = 911.6 \ mm \ Hg * 568 \ mL[/tex]

We are solving for the initial volume, so we must isolate the variable V₁. It is being multiplied by 248.71 millimeters of mercury. The inverse operation of multiplication is division, so we divide both sides by 248.71 mm Hg.

[tex]\frac {248.71 \ mm \ Hg * V_1 } {248.71 \ mm \ Hg}= \frac{911.6 \ mm \ Hg * 568 \ mL }{248.71 \ mm \ Hg}[/tex]

[tex]V_1 = \frac{911.6 \ mm \ Hg * 568 \ mL }{248.71 \ mm \ Hg}[/tex]

The units of millimeters of mercury (mm Hg) cancel.

[tex]V_1 = \frac{911.6 * 568 \ mL }{248.71 }[/tex]

[tex]V_1 = \frac {517788.8} {248.71 } \ mL[/tex]

[tex]V_1 = 2081.897793 \ mL[/tex]

Let's round to the nearest whole number. The 8 in the tenths place tells us to round the 1 up to a 2.

[tex]V_1= 2082 \ mL[/tex]

The gas initially occupied a volume of 2082 milliliters and choice A is correct.


Related Questions

What is the maximum most that can be extracted from 76 g og Cr2O3.
A) 36 g
B) 52 g
C) 104 g
D) 152 g

Answers

Answer:

D) 152 g

Explanation:

a mass of 1 mol Cr2O3 = 2 × 52 + 3 × 16 = 152 g number of moles of Cr2O3.

How many grams of Br are in 445 g CaBr2 ?

Answers

Answer:

156 g Br. Explanation: In order to figure out how many grams of bromine you get in that many grams of calcium bromide, CaBr2 , you must find ...

Answer:

156 g

Explanation:156 g

plss muje brilliant bana lo plss

Which of the following is a half-reaction? A. Zn+CuSO4−> B. 2Cl−−>Cl2+2e− C. H2+1/2O2−>H2O D. −>Cu+ZnSO4

Answers

Answer:

2Cl——>Cl2+2e-

Explanation:

It shows an electron loss or gain

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution before the addition of any HNO3. The Kb of NH3 is 1.8 × 10-5.

Answers

Answer:

[tex]pH=11.12[/tex]

Explanation:

Hello,

In this case, ammonia dissociation is:

[tex]NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)[/tex]

So the equilibrium expression:

[tex]Kb=\frac{[NH_4^+][OH^-]}{[NH_3]}[/tex]

That in terms of the reaction extent and the initial concentration of ammonia is written as:

[tex]1.8x10^{-5}=\frac{x*x}{0.10M-x}[/tex]

Thus, solving by using solver or quadratic equation we find:

[tex]x=0.00133M[/tex]

Which actually equals the concentration of hydroxyl ion, therefore the pOH is computed:

[tex]pOH=-log([OH^-])=-log(0.00133)=2.88[/tex]

And the pH from the pOH is:

[tex]pH=14-pOH=14-2.88\\\\pH=11.12[/tex]

Best regards.

A sample is found to contain 1.29×10-11 g of salt. Express this quantity in picograms

Answers

Answer:12.9e-12g or in short 12.9pg

Explanation:as p=1e-12

State what would be observed when the following pairs of reagents are mixed in a test tube.
C6H2COOH and Na2CO3(aq)
(ii) CH3CH2CH2OH and KMnO4 /H
(iii) CH3CH2OH and CH3COOH + conc. H2SO4 (iv) CH3CH = CHCH3 and Br2 /H2O​

Answers

Answer:

(i). C6H2COOH and Na2CO3(aq)

observation: Bubbles of a colourless gas (carbon dioxide gas)

(ii) CH3CH2CH2OH and KMnO4 /H

observation: The orange solution turns green.

[This is because oxidation of propanol to propanoic acid occurs]

(iii) CH3CH2OH and CH3COOH + conc. H2SO4

observation: A sweet fruity smell is formed.

[This is because an ester, diethylether is formed]

(iv) CH3CH = CHCH3 and Br2 /H2O

observation: a brown solution is formed.

.formular for Charles' law​

Answers

Explanation:

Hey there!

Charles law states that the volume of gas is directly proportional to the absolute temperature at constant pressure.

As the fact is, at a constant pressure the volume of fix amount of dry gas is directly proportional to absolute temperature from Charles law.

Since V and T are directly varying directly, we can write the equation using constant "k".

V/T = k

In this case the value of k depends on the pressure of gas, the amount of gas and also unit volume.

V/T = k .........(i)

Let us consider V1 and T1 the Volume and temperature of the ideal gases.

Then the equation is;

V1/T1 =k........(ii)

After this let change the temperature and volume be T2 and V2, respectively.

Then the equation is:

V2/T2 = k.......(iii)

Now; Equating equation (ii) and (iii)…

V1/T1 = V2/T2

So, this the formula. (i.e V1/T1 = V2/T2).

Hope it helps!

A 5-column table with 2 rows. Column 1 is labeled number of protons, with entries 20 and 9; column 2 is number of neutrons, with entries 20 and D; Column 3 is atomic number, with entries A and E; Column 4 is Mass Number, with entries B and 19, and Column 5 is Element (symbol) with entries C and F. Using the periodic table, complete the table to describe each atom. Type in your answers

Answers

Answer:

A    ⇒ 20

B    ⇒ 40

C    ⇒ Ca

D    ⇒ 10

E     ⇒ 9

F     ⇒ F

Explanation:

edge 2021

Answer:

the person above is correct

Explanation:

công thức phân tử của glucozo

Answers

C₆H₁₂O₆ is the molecular formula of gulcozo.

Alex drips hydrochloric acid onto a 5.9 g piece of magnesium in a single displacement reaction. How many molecules of HCl are required? 2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g)

how do you do this?

Answers

2.96 × 10²³ molecules is required if Alex drips hydrochloric acid onto a 5.9 g piece of magnesium in a single displacement reaction

According to this question, hydrochloric acid (HCl) reacts with magnesium metal (Mg) to produce magnesium chloride and hydrogen has as follows:

2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g)

Based on the balanced chemical equation above; 2 moles of HCl reacts with 1 mole of Mg.

Next, we convert 5.9g of Mg to moles by using the formula; mole = mass/molar mass

Where; Mg = 24g/mol

mole = 5.9/24

mole = 0.246mol of Mg

If 2 moles of HCl reacts with 1 mole of Mg. 0.246 mol of Mg will react with (0.246 × 2) = 0.492mol of HCl

To convert mole to number of molecules, we multiply number of moles by 6.02 × 10²³ (Avogadro's no.)

Hence, number of molecules of HCl required is 0.492mol × 6.02 × 10²³

= 2.96 × 10²³ molecules of HClHCl

Learn more: https://brainly.com/question/14948089

Flowers are vascular plants.
Vascular plants have
structures
that transport nutrients
and water within a plant.
A. ball-like
B. heart-like
C. box-like
D. tube-like

Answers

Answer:

I think tube like ... is it right

Answer:

The actual answer is flowers have heart like structures that transport nutrients and water with in the plants.

Calculate the pH of a buffer that is 0.13 M in lactic acid and 0.10 M in sodium lactate. Express your answer using two decimal places.

Answers

Answer:

pH of the buffer is 3.75

Explanation:

It is possible to find pH of a buffer using Hendersson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is molar concentration of the conjugate base and [HA] concentration of the weak acid

In the lactic buffer, pKa = 3.86. Lactic acid is the weak acid and its conjugate base is tha lactate salt. H-H equation for this buffer is:

pH = 3.86 + log [Lactate] / [Lactic acid]

Replacing with the concentrations of the problem:

pH = 3.86 + log [0.10M] / [0.13M]

pH = 3.75

pH of the buffer is 3.75

Rectangular cube 3.2 m length 1.2 m in height and 5 m in length is split into two parts. The container has a movable airtight divider that divides its length as necessary. Part A has 58 moles of gas and part B has 165 moles of a gas.

Required:
At what length will the divider to equilibrium?

Answers

Answer:

The length the divider is to  equilibrium from Part A = 1.30 m and from Part B = 3.70 m

Explanation:

Given that:

A rectangular cube with 3.2 m breadth, 1.2 m height and 5 m in length is splitted into two parts.

The diagrammatic expression for the above statement can be found in the attached diagram below.

The container has a movable airtight divider that divides its length as necessary.

Part A has 58 moles of gas

Part B has 165 moles of a gas.

Thus, the movable airtight divider will stop at a length where the pressure on it is equal on both sides.

i.e

[tex]\mathtt{P = P_A = P_B}[/tex]

Using the ideal gas equation,

PV = nRT

where, P,R,and  T are constant.

Then :

[tex]\mathsf{\dfrac{V_A}{n_A}= \dfrac{V_B}{n_B}}[/tex]

[tex]\mathsf{\dfrac{L_A \times B \times H}{n_A}= \dfrac{L_B \times B \times H}{n_B}}[/tex] --- (1)

since Volume of a cube = L × B × H

From the question; the L = 5m

i,e

[tex]\mathsf{L_A +L_B}[/tex] = 5

[tex]\mathsf{L_A = 5 - L_B}[/tex]

From equation (1) , we divide both sides by (B × H)

Then :

[tex]\mathsf{\dfrac{L_A }{n_A}= \dfrac{L_B }{n_B}}[/tex]

[tex]\mathsf{\dfrac{5-L_B}{58}= \dfrac{L_B }{165}}[/tex]

By cross multiplying; we have:

165 ( 5 - [tex]\mathsf{L_B}[/tex] )  = 58 (

825 - 165[tex]\mathsf{L_B}[/tex]  = 58

825 = 165[tex]\mathsf{L_B}[/tex] +58

825 = 223[tex]\mathsf{L_B}[/tex]

[tex]\mathsf{L_B}[/tex] = 825/223

[tex]\mathsf{L_B}[/tex]  = 3.70 m

[tex]\mathsf{L_A = 5 - L_B}[/tex]

[tex]\mathsf{L_A = 5 - 3.70}[/tex]

[tex]\mathsf{ L_A}[/tex] = 1.30 m

The length the divider is to  equilibrium from Part A = 1.30 m and from Part B = 3.70 m

A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
Br(g)
Cl2(g)
I2(g)
F2(g)
B. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2S(g)
H2O(g)
H2O2(g)
C. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous)
C(s, diamond)
C(s, graphite)

Answers

Answer:

A. Rank the following substances in order of decreasing standard molar entropy (S∘).

Rank the gases from largest to smallest standard molar entropy

I2(g)>Br2(g)>Cl2(g)>F2(g)

B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

H2O2(g)>H2S(g) >H2O(g)

C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

C(s, amorphous) >C(s, graphite)>C(s, diamond)

Explanation:

Hello,

In this case, we can apply the following principles to explain the order:

- The greater the molar mass, the larger the standard molar entropy.

- The greater the molar mass and the structural complexity, the larger the standard molar entropy.

- The greater the structural complexity, the larger the standard molar entropy.

A. Rank the following substances in order of decreasing standard molar entropy (S∘).

Rank the gases from largest to smallest standard molar entropy

I2(g)>Br2(g)>Cl2(g)>F2(g)

This is due to the fact that the greater the molar mass, the larger the standard molar entropy.

B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

H2O2(g)>H2S(g) >H2O(g)

This is due to the fact that the greater the molar mass and the structural complexity, the larger the standard molar entropy as the hydrogen peroxide has four bonds and weights 34 g/mol as well as hydrogen sulfide that has two bonds only.

C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

C(s, amorphous) >C(s, graphite)>C(s, diamond)

Since the molecular complexity is greater in the amorphous carbon (messy arrangement), mid in the graphite and lower in the diamond (well organized).

Regards.

Most reactions, including enzyme-catalyzed reactions, proceed faster at higher temperatures. However, for a given enzyme, the rate drops off abruptly at a certain temperature. Account for this behavior.

Answers

Answer:

It has denatured

Explanation:

When the temperature get high the enzymes tend to change shape and denaturing occurs.

ort
Which is a primary alcohol?
0 3-pentanol
2-propanol
1-ethanol
4-octanol
urvey
Lig A Moving to another question will save this response.

Answers

Answer:

1 ethanol is right answer

Explanation:

CH3- CH2-OH

For the following reaction, 3.76 grams of iron are mixed with excess oxygen gas . The reaction yields 4.29 grams of iron(II) oxide . iron ( s ) oxygen ( g ) iron(II) oxide ( s ) What is the theoretical yield of iron(II) oxide

Answers

Answer:

4.84g of FeO is the theoretical yield

Explanation:

The Iron, Fe(s), reacts with oxygen, O₂(g), producing Iron (II) oxide, as follows:

2Fe(s) + O₂(g) → 2FeO

Theoretical yield is the yield of a reaction in which you assume the 100% of reactants is converted in products.

To find theoretical yield we need to find moles of Iron, and, knowing 2 moles of Fe produce 2 moles of FeO (Ratio 1:1), we can find theoretical yield of FeO as follows:

Moles Fe (Molar mass: 55.845g/mol)

Using the molar mass of the compound we can convert grams to moles, thus:

3.76g Fe × (1mol / 55.845g) = 0.0673 moles of Fe

Moles and mass of FeO

As there are in reaction 0.0673 moles Fe, assuming a theoretical yield (And as ratio of the reaction is 1:1), you will obtain 0.0673 moles of FeO.

Theoretical yield is given in grams, As molar mass of FeO is 71.844g/mol, theoretical yield of the reaction is:

0.0673 moles FeO × (71.844g / mol) =

4.84g of FeO is the theoretical yield

How many valence electrons must two atoms share to form a single covalent bond? answers A.2 B.4 C.3 D.1

Answers

Answer:

2

Explanation:

A single covalent bond is formed when two electrons are shared between the same two atoms, one electron from each atom.

Answer:

the answer is 2

Explanation:

Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)
Find the volume if the density is 2.6 g/mL and the mass is 9.7 g.(5 pts)
Find the mass if the density is 1.6 g/cm3 and the volume is 4.1 cm3 (5 pts)
Find the density if the initial volume of water is 12.8 mL, the final volume is 24.6 mL and the mass of the object is 4.3 g. Make a drawing to show the water displacement using a graduated cylinder. (gdoc, gdraw)

Answers

Answer:

[tex]\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}[/tex]

Explanation:

1. Density from mass and volume

[tex]\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}[/tex]

2. Volume from density and mass

[tex]V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}[/tex]

3. Mass from density and volume

[tex]\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}[/tex]

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water                = 12.8 mL

Volume of object               = 11.8 mL

[tex]\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}[/tex]

Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

 

A baseball has a mass of 0.145 kilograms. If acceration due to gravity is 9.8m/s,what is the weight of the baseball in newtons?

Answers

Answer:

I hope it works

Explanation:

As we know that

w=m*g

given m=0.145 , g=9.8

hence we get

w= (9.8)*(0.145)

w=1.421 m/sec 2

if its help-full thank hit the stars and brain-list it thank you

ΔH = +572 kJ for the decomposition of water by the reaction 2 H 2O( ) → 2 H 2(g) + O 2(g). How many grams of water can be decomposed by 5.00 × 10 3 kJ of energy? ΔH = +572 kJ for the decomposition of water by the reaction 2 H 2O( ) → 2 H 2(g) + O 2(g). How many grams of water can be decomposed by 5.00 × 10 3 kJ of energy?

Answers

Answer:

315 g

Explanation:

Step 1: Write the thermochemical equation

2 H₂O(l) → 2 H₂(g) + O₂(g)      ΔH = +572 kJ

Step 2: Calculate the molar of water decomposed by 5.00 × 10³ kJ of energy

According to the thermochemical equation, 572 kJ are required to decompose 2 moles of water.

5.00 × 10³ kJ × (2 mol/572 kJ) = 17.5 mol

Step 3: Calculate the mass corresponding to 17.5 moles of water

The molar mass of water is 18.02 g/mol.

17.5 mol × 18.02 g/mol = 315 g

Determine the oxidation state for each of the elements below. The oxidation state of ... silver ... in ... silver oxide Ag2O ... is ... ___ . The oxidation state of sulfur in sulfur dioxide SO2 is ___ . The oxidation state of iron in iron(

Answers

Answer:

The oxidation state of silver in [tex]\rm Ag_2O[/tex] is [tex]+1[/tex].

The oxidation state of sulfur in [tex]\rm SO_2[/tex] is [tex]+4[/tex].

Explanation:

The oxidation states of atoms in a compound should add up to zero.

Ag₂O

There are two silver [tex]\rm Ag[/tex] atoms and one oxygen [tex]\rm O[/tex] atom in one formula unit of [tex]\rm Ag_2O[/tex]. Therefore:

[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].

The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is [tex]-2[/tex]. Silver oxide [tex]\rm Ag_2O[/tex] isn't an exception. Therefore:

[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}[/tex].

Solve this equation for the (average) oxidation state of [tex]\rm Ag[/tex]:

[tex]\text{Oxidation state of $\rm Ag$} = 1[/tex].

SO₂

Similarly, because there are one sulfur [tex]\rm S[/tex] atom and two oxygen [tex]\rm O[/tex] atoms in each [tex]\rm SO_2[/tex] molecules:

[tex]\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].

The oxidation state of [tex]\rm O[/tex] in [tex]\rm SO_2[/tex] is also [tex]-2[/tex], not an exception, either.

Therefore:

[tex]\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}[/tex].

Solve this equation for the oxidation state of [tex]\rm S[/tex] here:

[tex]\text{Oxidation state of $\rm S$} = 4[/tex].

Ozone (O 3) in the atmosphere can react with nitric oxide (NO): O 3(g) + NO(g) → NO 2(g) + O 2(g). Calculate the ΔG° for this reaction at 25°C. (ΔH° = –199 kJ/mol, ΔS° = –4.1 J/K·mol)

Answers

Answer:

ΔG°  = 1022. 8 kJ

Explanation:

ΔH° = –199 kJ/mol

ΔS° = –4.1 J/K·mol

T = 25°C = 25 + 273 = 298K (Converting to kelvin temperature)

ΔG° = ?

The relationship between these varriables are;

ΔG° = ΔH°  - TΔS°

ΔG° = –199 - 298 (–4.1)

ΔG° = -199 + 1221.8

ΔG°  = 1022. 8 kJ

To calculate ΔG, the following equation should be used:

ΔG° = ΔH°  - TΔS°

Given:

ΔH° = –199 kJ/mol,  

ΔS° = –4.1 J/K·mol

T=25+273K=298K  

Substitute the respective values:

ΔG° = ΔH°  - TΔS°

       =–199 kJ/mol-(298K×(-4.1 J/K·mol*(1KJ/1000J)))

       =-197.78kJ/mol

Thus, we can conclude the value of ΔG°=-197.78kJ/mol  

Learn more about ΔG° here:

https://brainly.com/question/2491198

Given the following balanced reaction: 2Na(s) + F2(g) --> 2NaF(s)
a) How many moles of NaF will be made from 2.6 moles of F2?

b) How many moles of NaF will be made from 4.8 moles of Na?

Answers

Answer:

yes it is corrwect iyt is absolitle correct

Explanation:

Please help me! I am a bit stuck on this.

Answers

Option 2 is the correct answer

What is the primary source of energy in most living communities?

Answers

Answer:

The sun

Explanation:

The sun is the primary source of energy in most living communities. The producers or the green plants that prepare their own food by the use of sunlight and other natural resources. Carbon dioxide, water, and other minerals are used by the plants to make their food in the presence of chlorophyll. Plants are then consumed by the consumers. This chain helps in forming the food chain and the food web.

A student carries out the precipitation reaction shown below, starting with 0.030 moles of calcium nitrate. The final mass of the precipitate is 2.9 g. Answer the questions below to determine the percent yield. 3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq) 1. a. Which product is the precipitate? b. How many moles of the precipitate would one expect to be produced from 0.030 moles of calcium nitrate? c. How many grams of solid do you expect to be produced? d. What is the percent yield?

Answers

Answer:

a. Ca₃(PO₄)₂.

b. 0.010 moles of Ca₃(PO₄)₂ can we expect to be produced

c. 3.1g of Ca₃(PO₄)₂

d. Percent yield = 93.5%

Explanation:

a. Based on the reaction:

3Ca(NO₃)₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6NaNO₃(aq)

3 moles of calcium nitrate reacts with 2 moles of sodium phosphate producieng 1 mole of calcium phosphate.

As you can see, Ca₃(PO₄)₂ is a solid product -(s)-, that means when the reaction occurs the precipitate produced is the solid,

Ca₃(PO₄)₂

b. As 3 moles of calcium nitrate produce 1 mole of calcium phosphate and there are 0.030 moles of calcium nitrate

0.030 moles Ca(NO₃)₂ × (1 mol Ca₃(PO₄)₂ / 3 moles Ca(NO₃)₂) =

0.010 moles of Ca₃(PO₄)₂ can we expect to be produced

c. As molar mass of Ca₃(PO₄)₂ is 310.18g/mol, the mass of 0.010 moles (The expected mass) is;

0.010 moles Ca₃(PO₄)₂ × (310.18g / mol) =

3.1g of Ca₃(PO₄)₂

d. The percent yield is defined as 100 times the ratio between the obtained yield (That is 2.9g of precipitate, Ca₃(PO₄)₂) and the expected yield, 3.1g of Ca₃(PO₄)₂:

[tex]\frac{2.9g}{3.1g} *100[/tex]

Percent yield = 93.5%

(a) The product in solid state would be the precipitate. Hence, the precipitate would be Ca3(PO4)2

(b) From the balanced equation of the reaction: 3 moles of Ca(NO3)2 is required for 1 mole of Ca3(PO4)2

If there are just 0.030 moles of Ca(NO3)2, then"

3 moles = 1

0.030 moles =    1 x 0.030/3

                         = 0.01 moles of Ca3(PO4)2

In other words, 0.01 moles of the precipitate would be expected to be produced from 0.030 moles of calcium nitrate.

(c) 0.01 moles solid (Ca3(PO4)2) is expected. Mass of Ca3(PO4)2 expected:

      mass   = mole x molar mass

molar mass of Ca3(PO4)2 = 310.18 g/mol

mass of Ca3(PO4)2 expected to be produced = 0.01 x 310.18

                                                                       = 3.1018 g

Hence, 3.1018g of solid is expected to be produced.

(d) Percentage yield = actual yield/theoretical yield x 100

                          = 2.9/3.1018 x 100

                               = 93.5%

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Provide the name of the compound shown below. Spelling and format counts!
CH3
NO2
NO2​

Answers

Answer:

ch3:-methanide

No2:-nitrogen dioxide

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What is the Equation of Reduction in Mg+F2 gives MgF2, I WILL MARK YOU AS BRAINLIST

Answers

Answer:

Mg+F2= Mgf2

Explanation:

F 2 is an oxidizing agent, Mg is a reducing agent. ; Pale-yellow to greenish gas with a pungent, irritating odor.

Give 3 examples of chemical change and physical change in daily life

Answers

Explanation:

Physical change

★ Boiling of water

❤︎ freezing of water

❣︎ washing of clothes

♫︎ mixing black and brown stones

Chemical change

➪ making of dough

㋛︎ burning of wood

❥︎ cooking of the raw food

❁ burning of paper

ꨄ︎ rusting of iron

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