Does the addition of a catalyst affect the ΔH of reaction?

Answer:

1.932 kg

Explanation:

First we convert 100.0 mL to L:

Then we calculate the mass of the substance, using the definition of density:

As the multiplication involves two numbers of 4 significant figures each, the answer needs to have 4 significants figures as well.

above because its above

Answer:

10573.9K

Explanation:

Using pressure law equation;

P1/T1 = P2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

P1 = 46mmHg

P2 = 760mmHg

T1 = 640K

T2 = ?

Using P1/T1 = P2/T2

46/640 = 760/T2

Cross multiply

640 × 760 = 46 × T2

486400 = 46T2

T2 = 486400 ÷ 46

T2 = 10573.9K

Answer:

9460 kj

Explanation:

just did it

Answer: 63.96%.

Explanation:

In 111g of Calcium chloride, there is 40g of Calcium and 71g of Chlorine. Percentage Composition of Chlorine is 63.96%.

Hope This Helps!

Answer:

Explanation:

mass = 400 grams * [1 kg/1000 grams] = 0.400 kg

c = 387 Joules / (oC * kg)

Δt = 55 - 30 = 25 oC

E = m*c * Δt

E = 0.4 * 387 * 25

E = 3870 Joules

Answer:

Number of moles  of sodium dissolved =  6.0 *10^23

Explanation:

The image for the question is attached

Solution

a) Total 181 ions of Na are dissolved

b)

The number of moles of sodium dissolved = 181/6.023 *10^23

Number of moles  of sodium dissolved = 5.987 * 10^23

Number of moles  of sodium dissolved =  6.0 *10^23

Answer:

Ovary

Explanation:

ovaries produce the most estrogen in females.

Answer:

I say the correct answers are primary and secondary and teriary.

Explanation:

I say you are right!!

[tex]\huge\mathsf{\red{\underline{\underline{Compound}}}}[/tex]

[tex]{\green{\dashrightarrow}}[/tex]A chemical compound is a chemical substance that is made of two or more atoms of different elements that share a chemical bond.

[tex]{\green{\dashrightarrow}}[/tex]A chemical formula represents the ratio of atoms per element that make up the chemical compound.

[tex]\large{\pink{\sf{5~ Examples~ of~ Compound~ are:-}}}[/tex]

Answer:

1. Empirical formula => C₂H₃O

2. Molecular formula => C₆H₉O₃

Explanation:

From the question given above, the following data were obtained:

Mass of compound = 50 g

Mass of Carbon = 24.66 g

Mass of Hydrogen = 3.43 g

Molecular weight of compound = 146.0 amu

Empirical formula =?

Molecular formula =?

Next, we shall determine the mass of oxygen in the compound. This can be obtained as follow:

Mass of compound = 50 g

Mass of C = 24.66 g

Mass of H = 3.43 g

Mass of O =?

Mass of O = mass of compound – ( mass of C + mass of H)

= 50 – (24.66 + 3.43)

= 50 – 28.09

= 21.91 g

1. Determination of the empirical formula.

Mass of C = 24.66 g

Mass of H = 3.43 g

Mass of O = 21.91 g

Divide by their molar mass

C = 24.66 / 12 = 2.055

H = 3.43 / 1 = 3.43

O = 21.91 / 16 = 1.369

Divide by the smallest

C = 2.055 / 1.369 = 2

H = 3.43 / 1.369 = 3

O = 1.369 / 1.369 = 1

Therefore, the empirical formula of the compound is C₂H₃O

2. Determination of the molecular formula.

Molecular weight of compound = 146.0 amu

Empirical formula => C₂H₃O

Molecular formula =?

Molecular formula = [C₂H₃O]ₙ = molecular weight

Thus,

[C₂H₃O]ₙ = 146

[(12×2) + (3×1) + 16]n = 146

[24 + 3 + 16]n = 146

43n = 146

Divide both side by 43

n = 146 / 43

n = 3

Molecular formula = [C₂H₃O]ₙ

Molecular formula = [C₂H₃O]₃

Molecular formula = C₆H₉O₃

Answer:

In order from left to right, 7 (gamma), 5 (ultraviolet, now continue pattern), 4, 6, 2, 3, 1.

Answer:

Increasing its mass to twice its original value

Explanation:

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Answer:

I think it's 1.D

2.C

3.C

my apologies if it's incorrect

Answer: There are [tex]16.14 \times 10^{23}[/tex] atoms of hydrogen are present in 40g of urea, [tex](NH_{2})_{2}CO[/tex].

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol[/tex]

According to the mole concept, 1 mole of every substance contains [tex]6.022 \times 10^{23}[/tex] atoms.

So, the number of atoms present in 0.67 moles are as follows.

[tex]0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms[/tex]

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

[tex]4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms[/tex]

Thus, we can conclude that there are [tex]16.14 \times 10^{23}[/tex] atoms of hydrogen are present in 40g of urea, [tex](NH_{2})_{2}CO[/tex].

Answer: The molar concentration of 29 g of [tex]Mg(OH)_{2}[/tex] dissolved in 1.00 L of water is 0.497 M.

Explanation:

Given: Mass = 29 g

Volume = 1.00 L

Moles is the mass of substance divided by its molar mass. So, moles of [tex]Mg(OH)_{2}[/tex] is as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{29 g}{58.32 g/mol}\\= 0.497 mol[/tex]

Molarity is the number of moles of a substance divided by volume in liter.

Hence, molarity of the given solution is as follows.

[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.497 mol}{1.00 L}\\= 0.497 M[/tex]

Thus, we can conclude that the molar concentration of 29 g of [tex]Mg(OH)_{2}[/tex] dissolved in 1.00 L of water is 0.497 M.

Answer:

Molarity = 0.809 M

mole fraction = 0.047

Explanation:

The complete question is

Calculate the molarity and mole fraction of acetone in a 1.09-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.

Solution -

Solution for molarity:

1.09-molal means 1.09 moles of acetone in 1.00 kilogram of ethanol.

1)  

Mass of 1.09 mole of acetone

= 1.09  mol x 58.0794 g/mol = 63.306 g

Density of acetone = 0.788 g/cm3  

Thus, volume of 1.09 moles of acetone = 63.306 g/0.788 g/cm3 = 80.34 cm3

For ethanol

1000 g divided by 0.789 g/cm3 = 1267.427 cm3

Total volume of the solution = Volume of acetone + Volume of ethanol = 80.34 cm3 + 1267.427 cm3 = 1347.765 cm3  = 1.347 L

a) Molarity:

1.09 mol / 1.347 L = 0.809 M

Mole Fraction  

a) moles of ethanol:

1000 g / 46.0684 g/mol = 21.71 mol

b) moles of acetone:

1.09 / (1.09 + 21.71) = 0.047

Answer:

NaOH is the limiting reactant.

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary to write out the chemical reaction between NaOH and HCl:

[tex]NaOH+HCl\rightarrow NaCl+H_2O[/tex]

Thus, since they react in a 1:1 mole ratio; we can now calculate the moles of each substance by using their volumes and molarities:

[tex]n_{NaOH}=0.0250L*0.81mol/L=0.02025molNaOH\\\\n_{HCl}=0.0650L*0.33mol/L=0.02145molHCl[/tex]

Now, since NaOH is in a fewer proportion, we infer just 0.02025 moles of HCl are consumed so that 0.0012 moles of this acid remain unreacted; in such a way, we infer that the NaOH is the limiting reactant for this reaction.

Regards!

Answer:

2Na and 2HCl

Explanation:

The equation for the above chemical reaction is as follows:

2Na + 2HCI → 2NaCl + H2

In a reaction, the reactants are said to be those that combine together to form products. In this case, sodium (Na) and hydrochloric acid (HCl) are the reactants of this reaction

Answer:

The correct answer is option 4, that is, there is exactly enough sodium carbonate.

Explanation:

Based on the given question, the reaction will be,

2 HCl (aq) + Na2CO3 (s) ⇒ 2 NaCl (aq) + CO2 (g) + H2O (l)

Therefore, for neutralizing 2 moles of HCl, one mole of Na2CO3 is required.

No of moles present in 1 Kg or 1000 grams of Na2CO3 will be,

Moles = Weight/Molecular mass of Na2CO3

Moles = 1000 / 106 = 9.43

Thus, 9.43 moles of Na2CO3 is present.

No of moles present in 1 liter of 9.75 M HCl is 9.75.

No. of moles present in 5 Liters of HCl (9.75 M),

= 5 × 9.75 = 48.75

Thus, for 2 moles of HCl 1 mole of Na2CO3 is required. Now for 48.75 moles of HCl, the moles required of Na2CO3 is 9.75. Therefore, for complete neutralization, the moles of Na2CO3 required is 9.75, and the present moles is 9.43.

Hence, there is exactly enough sodium carbonate.

Answer:

Hello, I was doing this and from my caluclations it should be :

Explanation:

3 Moles.

The three moles of MgCO3 are present in 252.939 grams of MgCO3.

So, option B is correct one .

The molar mass of substance is equal to mass of one mole of  that substance.

Example: Molar mass of MgCO3 is equal to mass of one mole of MgCO3.

Molar mass of Mg = 24.305 u

Molar mass of C = 12.011 u

Molar mass of O = 15.999 u

Molar mass of MgCO3 = Molar mass of Mg + Molar mass of C + Molar mass of O*3

Molar mass of MgCO3 = 24.305 + 12.011  +  15.999 * 3

Molar mass of MgCO3 = 84.3139 gram

Molar mass of  MgCO3 contain  84.3139 gram

So, 252.939 grams of MgCO3 = 252.939 grams/84.3139 gram

252.939 grams of MgCO3 = 2.999 moles

To learn more about moles here.

https://brainly.com/question/26416088

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Answer:

[tex]\boxed {\boxed {\sf 0.255 \ mol \ C }}[/tex]

Explanation:

If we want to convert from grams to moles, the molar mass is used. This is the mass of 1 mole. They are found on the Periodic Table as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

Look up the molar mass of carbon.

Set up a ratio using the molar mass.

[tex]\frac {12.011 \ g \ C}{ 1 \ mol \ C}[/tex]

Since we are converting 3.06 grams to moles, we multiply by that value.

[tex]3.06 \ g \ C*\frac {12.011 \ g \ C}{ 1 \ mol \ C}[/tex]

Flip the ratio. This way, the ratio is still equivalent, but the units of grams of carbon cancel.

[tex]3.06 \ g \ C* \frac{1 \ mol \ C}{12.011 \ g\ C}[/tex]                      

[tex]3.06 * \frac{1 \ mol \ C}{12.011 }[/tex]    

[tex]\frac {3.06}{12.011 } \ mol \ C[/tex]                                

[tex]0.25476646 \ mol \ C[/tex]

The original measurement of grams (3.06) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

The 7 in the ten-thousandth place tells us to round the 4 up to a 5.

[tex]0.255 \ mol \ C[/tex]

3.06 grams of carbon is approximately 0.255 moles of carbon.

Answer:

Following are the solution to the given question:

Explanation:

Please find the matching in attached file.

During the electrolysis of Molten LiBr : Li is reduced and Br are oxidized .

Lithium Metal is produced at the Cathod during the electrolysis of Molten LiBR .

In the aquous LiBr : In aquous LiBr potential of Li is greater than the of water then Li is reduced to produce solid Li ion.  As well As aqueous LiF ( electrolysis)and Molten LiF electrolysis produce the gas.  

aqueous LiF: Oxygen gas.

Molten LiF : Flourine gas.  

Answer: look at the explanation and try to work it

Explanation: in contrast, our atmosphere blocks most ultraviolet light (UV) and all X-rays and gamma-rays from reaching the surface of Earth. Because of this, astronomers can only study these kinds of light using detectors mounted on weather balloons, in rockets, or in Earth-orbiting satellites.

Explanation:

H=mc×∆©

9.86=m×0.902×(30.5-23.2)

m=1.5

Explanation:

The specific heat of a substance is the amount of heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius. The formula for calculating the heat absorbed or released by a substance is `q = mcΔT`, where `q` is the heat absorbed or released, `m` is the mass of the substance, `c` is the specific heat of the substance, and `ΔT` is the change in temperature.

In this case, we can use this formula to solve for the mass of the aluminum sample. We know that `q = 9.86 J`, `c = 0.902 J/g°C`, and `ΔT = 30.5°C - 23.2°C = 7.3°C`. Plugging these values into the formula, we get:

`9.86 J = m * 0.902 J/g°C * 7.3°C`

Solving for `m`, we find that the mass of the aluminum sample is approximately `1.5 g`, rounded to 2 significant figures.

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Answer:

The amount after three year is 617934.1302

Explanation:

Complete question

A person places $530,000 in an investment account earning an annual rate of 5.25%, compounded continuously. Using the formula V = Pe^{rt}V=Pe rt , where V is the value of the account in t years, P is the principal initially invested, e is the base of a natural logarithm, and r is the rate of interest, determine the amount of money, to the nearest cent, in the account after 3 years

Solution

The formula for calculating compound interest is

[tex]A = p (1 + \frac{r}{n})^{nt}[/tex]

Substituting the given values we get -

[tex]A = 530,000 (1 + \frac{5.25}{100})^3\\A = 530,000 * ( 1+ 0.0525)^3\\A = 530,000 * ( 1.0525)^3\\A = 617934.1302[/tex]

The amount after three year is 617934.1302

Answer:

The heat capacity of the metal underneath the gold is 0.431 J/g°C

Explanation:

Using the formula as outlined in the image:

Q = m × c × ∆T

Where;

Q = amount of heat energy (J)

m = mass of substance (g)

c = specific heat capacity (J/g°C)

∆T = change in temperature (°C)

According to the information in this question;

Q = 503.9J

m = 23.02g

c = ?

∆T = 74°C - 23.2°C = 50.8°C

Using Q = m × c × ∆T

c = Q ÷ m∆T

c = 503.9 ÷ (23.02 × 50.8)

c = 503.9 ÷ 1169.42

c = 0.431 J/g°C

From the above heat capacity of the metal underneath the gold, it is obvious that the metal is not pure gold (c = 0.129J/g°C)

Answer:

I think it is 7.60 X 1

but if it's not srry

Answer:

lololol

Explanation: