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kirza4 [7]
3 years ago
5

What is the mass of a sample of NH3 containing 6.3 × 1024 molecules of NH3?

Chemistry
1 answer:
quester [9]3 years ago
7 0

Answer:

178 grams

Explanation:

<em>It is known that 1.0 mole of a compound contains Avogadro's number of molecules (6.022 x 10²³).</em>

<em><u>Using cross multiplication:</u></em>

1.0 mol contains → 6.022 x 10²³ molecules.

??? mol contains → 6.3 x 10²⁴ molecules.

∴ The no. of moles of (6.3 x 10²⁴ molecules) of NH₃ = (1.0 mol)(6.3 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 10.46 mol.

<em>∴ The no. of grams of NH₃ present = no. of moles x molar mass</em> = (10.46 mol)(17.0 g/mol) = <em>177.8 g ≅ 178 g.</em>

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Explanation:

Let us assume that the given data is as follows.

   mass of barium acetate = 2.19 g

   volume = 150 ml = 0.150 L    (as 1 L = 1000 ml)

   concentration of the aqueous solution = 0.10 M

Therefore, the reaction equation will be as follows.

        Ba(C_{2}H_{3}O_{2})_{2} \rightarrow Ba^{2+} + 2C_{2}H_{3}O^{-}_{2}

Hence, moles of C_{2}H_{3}O^{-}_{2} = 2 \times Ba(C_{2}H_{3}O_{2})_{2}  .......... (1)

As,   No. of moles = \frac{mass}{\text{molar mass}}

Hence, moles of Ba(C_{2}H_{3}O_{2})_{2} will be calculated as follows.                          

     No. of moles = \frac{mass}{\text{molar mass}}  

                          =  \frac{2.19 g}{255.415 g/mol}   (molar mass of Ba(C_{2}H_{3}O_{2})_{2} is 255.415 g/mol)            

                       = 8.57 \times 10^{-3}

    Moles of C_{2}H_{3}O^{-}_{2} = 2 \times 8.57 \times 10^{-3}

                          = 0.01715 mol

Hence, final molarity will be as follows.

              Molarity = \frac{\text{no. of moles}}{volume}

                             = \frac{0.01715 mol}{0.150 L}

                             = 0.114 M

Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.

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Answer:

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Explanation:

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A. At STP, what is the volume of 708 mol of nitrogen gas? 708 mol = 708 mol X L B. A sample of hydrogen gas occupies 14.1 L at S
elena-14-01-66 [18.8K]

Answer:

A. 15859.2 L or 15900 L

B. 0.629 mol

Explanation:

At STP, one mole is equal to approximately 22.4 L

L or mL is volume, so you are attempting to solve for L or mL.

A.

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B.

(14.1 L) x (1 mole/ 22.4 L) = 0.629 mol.

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Micheal has a substints that he puts in container 1 the substance has a volume of 5 cubic meters
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