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3 years ago

Helpppp Show using two conversion factors how you would convert from 0.020 kg into mg.

Chemistry

1 answer:

MissTica3 years ago

7 0

Answer:

20000 mg

Explanation:

1 kg = 1000000

so if u multiply 0.020kg by 1000000 u get 20000

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A student needs to make a solution of 308 mL of 0.02 M of Copper (II) Sulfate. All they have available is a solution of 0.250 M.

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Answer:

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3 years ago

A vessel of volume 22.4 dm3 contains 20 mol h2 and 1 mol n2 ad 273.15 k initially. All of the nitrogen reacted with sufficient h

NikAS [45]

Nitrogen combine with hydrogen to produce ammonia $\text{NH}_3$ at a $1:3:2$ ratio:

$\text{N}_2 \; (g) + 3 \; \text{H}_2 \; (g) \leftrightharpoons 2\; \text{NH}_3 \; (g)$

Assuming that the reaction has indeed proceeded to completion- with all nitrogen used up as the question has indicated. $3 \; \text{mol}$ of hydrogen gas would have been consumed while $2 \; \text{mol}$ of ammonia would have been produced. The final mixture would therefore contain

$17 \; \text{mol}$ of $\text{H}_2 \; (g)$ and
$2 \; \text{mol}$ of $\text{NH}_3 \; (g)$

Apply the ideal gas law to find the total pressure inside the container and the respective partial pressure of hydrogen and ammonia:

$\begin{array}{lll} P(\text{container}) &= & n \cdot R \cdot T / V \\ & = & (17 + 2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 1.926 \times 10^{3} \; \text{kPa} \end{array}$
$\begin{array}{lll} P(\text{H}_2) &= & n \cdot R \cdot T / V \\ & = & (17) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 1.723 \times 10^{3} \; \text{kPa} \end{array}$
$\begin{array}{lll} P(\text{NH}_3) &= & n \cdot R \cdot T / V \\ & = & (2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 2.037 \times 10^{2} \; \text{kPa} \end{array}$

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3 years ago

Neutralization Reactions

PolarNik [594]

Answer:

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3 years ago

What is difference between pie and cake

Leokris [45]

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3 years ago

Ammonium chloride, ammonium sulfate and ammonium nitrate are used in fertilisers.

Hoochie [10]

Answer:

1.0625 g

Explanation:

Number of moles in 5g of NH4NO3 = 5.0g/80g/mol = 0.0625 moles

Again the balanced reaction equation is;

NH3 + HNO3 → NH4NO3

So,

1 mole of NH3 yields 1 mole of NH4NO3

x moles of NH3 yields 0.0625 moles of NH4NO3

x = 0.0625 moles of NH3

Mass of NH3 = number of moles * molar mass

Mass of NH3 = 0.0625 moles * 17g/mol

Mass of NH3 = 1.0625 g

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3 years ago