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3 years ago

Calculate the average atomic mass for the following elements.

Chemistry

1 answer:

nikklg [1K]3 years ago

8 0

Explanation:

All boron isotopes have 5 protons.

B11 has 6 neutrons, so its atomic mass is:

6 (1.008664 u) + 5 (1.007276 u) = 11.0884 u

B10 has 5 neutrons, so its atomic mass is:

5 (1.008664 u) + 5 (1.007276 u) = 10.0797 u

So the average atomic mass of boron is:

0.80 (11.0884 u) + 0.20 (10.0797 u) = 10.8866 u

The closest whole number is 11.

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What is the final volume?

zaharov [31]

Answer:

Option A. 9.4 L

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 8 L

Initial temperature (T₁) = 293 K

Final temperature (T₂) = 343 K

Final volume (V₂) =?

V₁ / T₁ = V₂ / T₂

8 / 293 = V₂ / 343

Cross multiply

293 × V₂ = 8 × 343

293 × V₂ = 2744

Divide both side by 293

V₂ = 2744 / 293

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3 years ago

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Answer:

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Explanation:

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2 years ago

Which of the following is the correct formula used for calculating Speed?

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3 years ago

How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci

bezimeni [28]

Answer:

Approximately $0.08\; \rm mol$ , assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure: $25\; \rm ^{\circ}C$ and $P = 101.325 \; \rm kPa$ .

The question states that the volume of this gas is $V = 2\; \rm dm^{3}$ .

Convert the unit of all three measures to standard units:

$\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}$ .

$\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}$ .

$\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}$ .

Look up the ideal gas constant in the corresponding units: $R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}$ .

Let $n$ denote the number of moles of this gas in that $V = 2\; \rm dm^{3}$ . By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

$P \cdot V = n \cdot R \cdot T$ .

Rearrange this equation and solve for $n$ :

$\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}$ .

In other words, there is approximately $2\; \rm mol$ of this gas in that $V = 2\; \rm dm^{3}$ .

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3 years ago