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2 years ago

Convert 7.83×1024 atoms of carbon to moles of carbon

Chemistry

2 answers:

mr_godi [17]2 years ago

6 0

Answer:

13.0 mol

Explanation:

Use Avogadro’s number to convert atoms of C to moles of C.

1 mol C atoms = 6.022 × 10²³ atoms of C

Moles of C = 7.83 ×10²⁴ × 1 /6.022 × 10²³

Moles of C = 13.0 mol C

SVETLANKA909090 [29]2 years ago

3 0

did you mean 10^24 atoms?

if so, youd do (7.83x10^24)/ (6.02x 10^23)---- avogadros constant

=13.01 moles

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What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a

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Answer:

$\boxed {\boxed {\sf 333 \ grams}}$

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

$Q= mc \Delta T$

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

ΔT= final temperature - initial temperature

The sample was heated from 58.8 degrees Celsius to 88.9 degrees Celsius.

ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

Q= 4500.0 J
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ΔT = 30.1 °C

Substitute these values into the formula.

$4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)$

Multiply on the right side of the equation. The units of degrees Celsius cancel.

$4500.0 \ J = m (13.52694 J/g)$

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

$\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}$

The units of Joules cancel.

$\frac {4500.0 \ J }{13.52694 J/g}= m$

$332.6694729 \ g =m$

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

$333 \ g \approx m$