What mineral is the hardest known substance in nature?

How many moles are in 1.2x10^3 grams of ammonia, NH3?

KonstantinChe [14]

I don't have a calculator with me right now, but that mass would be 1200 grams. Divide the given amount of grams by the molar mass of NH3, which is 17.031g/mol. (Nitrogen + 3(hydrogen)). Again, sorry I didn't have a calculator. But all you would need to do is divide 1200 by 17.031. If you need to use sig figs, your answer should have 2 because the 1.2 x 10^3 limits your amount of sig figs.

6 0

3 years ago

A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For NaOH:

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

For sulfuric acid:

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

For sodium sulfate:

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

For sulfuric acid:

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

For NaOH:

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0

3 years ago

How many miles of C atoms are present in a sample of 3.85x1024 C atoms?

alekssr [168]

Answer:
number of moles = 6.393 moles

Explanation:
One mole of any substance contains Avogadro's number (6.022 * 10^23) of atoms.
Therefore, to know the number of moles that contain 3.85 * 10^24 atoms, all we have to do is cross multiplication as follows:
1 mole ......................> 6.022 * 10^23
?? moles ..................> 3.85 * 10^24
number of moles = (3.85 * 10^24 *1) / (6.022 * 10^23)
number of moles = 6.393 moles

Hope this helps :)

8 0

3 years ago

5.50 L of a 0.190 M CaCl2M CaCl2 solution How do you find Grams?

Firdavs [7]

Answer:

115.891

Explanation:

Molarity is equal to moles/liters so to get moles we multiply molarity by liters, after this we multiply the moles by the total atomic weight of the compound to get grams

6 0

3 years ago

One application of Hess's Law (which works for ΔH, ΔS, and ΔG) is calculating the overall energy of a reaction using standard en

VikaD [51]

Answer:

The standard change in free energy for the reaction = - 437.5 kj/mole

Explanation:

The standard change in free energy for the reaction:

4 KClO₃ (s) → 3 KClO₄(s) + KCl(s)

Given that ΔGf(KClO3(s)) = -290.9 kJ/mol;

ΔGf(KClO4(s)) = -300.4 kJ/mol;

ΔGf(KCl(s)) = -409 kJ/mol

According to Hess's law

ΔGr (Free energy change of reaction)= ∑(Product free energy - reactant free energy)

⇒ ΔGr⁰ = {3 x (-300.4) + (-409)} - {3 x (- 290.9)}

= - 901.2 - 409 + 872.7

= - 437.5 kj/mole

3 0

3 years ago