What is the mass of a sample of material that has a volume of 6260 and a density of 6980 g/cm

Chemistry

1 answer:

svet-max [94.6K]3 years ago

8 0

The mass of the sample is 436.9 t.

Mass = 6260 cm³× (6980 g/1 cm³) = 4.369 × 10⁸ g = 4.369 × 10⁵ kg =436.9 t

You might be interested in

If 2 molecules of one reactant combine with 3 molecules of another to produce 5 molecules of a product, then what is the represe

zhenek [66]

Answer:- $2A+3B\rightarrow 5C$

Explanations:- Let's say the first molecule is A and the second molecule is B and the product molecule is C.

From given information, 2 molecules of A react with 3 molecules of B to produce 5 molecules of C.

We write the reactants on left side and the product on right side and there is an arrow between them from reactants towards the products.

So, the reaction could be represented as:

$2A+3B\rightarrow 5C$

3 0

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:

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