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kenny6666 [7]
3 years ago
15

Can anyone help with these ?

Chemistry
1 answer:
andriy [413]3 years ago
5 0
Uh no I can’t help but, I hope you have a good day!
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How many moles are in 1.2x10^3 grams of ammonia(NH3) ?
miskamm [114]

Answer : The number of moles present in ammonia is, 70.459 moles.

Solution : Given,

Mass of ammonia = 1.2\times 10^3g

Molar mass of ammonia = 17.031 g/mole

Formula used :

\text{Moles of }NH_3=\frac{\text{ given mass of }NH_3}{\text{ molar mass of }NH_3}

\text{Moles of }NH_3=\frac{1.2\times 10^3g}{17.031g/mole}=70.459moles

Therefore, the number of moles present in ammonia is, 70.459 moles.

5 0
3 years ago
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What is the mole fraction of acetic acid in a solution containing 2 moles of vinegar and 3 moles of water? A. 2 B. 3 C. 0.4 D. 0
9966 [12]

Answer:

C. 0.4.

Explanation:

<em>∵ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) = (no. of moles acetic acid)/(no. of moles of acetic acid + no. of moles of water).</em>

<em></em>

- no. of moles of acetic acid = 2, no. of moles of water = 3.

- Total no. of moles = no. of moles of acetic acid + no. of moles of water = 2 + 3 = 5.

<em>∴ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) =</em> (2)/(5)<em> = 0.4.</em>

3 0
3 years ago
Read 2 more answers
Which statements about exothermic and endothermic reactions are correct? (its a multiple choice question)
geniusboy [140]

Answer:

C)1 and 3 only☺️

Explanation:

5 0
3 years ago
Write the equation for changing atmospheric nitrogen and hydrogen to ammonia?
Pepsi [2]
NEED MORE INFORMATION
3 0
4 years ago
20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the
kirill115 [55]

Answer:

Choice A: approximately 0.12\; \rm M.

Explanation:

Note that the unit of concentration, \rm M, typically refers to moles per liter (that is: 1\; \rm M = 1\; \rm mol\cdot L^{-1}.)

On the other hand, the volume of the two solutions in this question are apparently given in \rm cm^3, which is the same as \rm mL (that is: 1\; \rm cm^{3} = 1\; \rm mL.) Convert the unit of volume to liters:

  • V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L.
  • V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L.

Calculate the number of moles of \rm H_2SO_4 formula units in that 0.02\; \rm L of the 0.09\; \rm M solution:

\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}.

Note that \rm H_2SO_4 (sulfuric acid) is a diprotic acid. When one mole of \rm H_2SO_4 completely dissolves in water, two moles of \rm H^{+} ions will be released.

On the other hand, \rm NaOH (sodium hydroxide) is a monoprotic base. When one mole of \rm NaOH formula units completely dissolve in water, only one mole of \rm OH^{-} ions will be released.

\rm H^{+} ions and \rm OH^{-} ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid \rm H_2SO_4 dissolves in water completely, it will take two moles of \rm OH^{-} to neutralize that two moles of \rm H^{+} produced. On the other hand, two moles formula units of the monoprotic base \rm NaOH will be required to produce that two moles of \rm OH^{-}. Therefore, \rm NaOH and \rm H_2SO_4 formula units would neutralize each other at a two-to-one ratio.

\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O.

\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2.

Previous calculations show that 0.0018\; \rm mol of \rm H_2SO_4 was produced. Calculate the number of moles of \rm NaOH formula units required to neutralize that

\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}.

Calculate the concentration of a 0.03\; \rm L solution that contains exactly 0.0036\; \rm mol of \rm NaOH formula units:

\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}.

3 0
3 years ago
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