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3 years ago

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Calculate a mass of 12.044×10^23 of calcium carbonate

1 answer:

tigry1 [53]3 years ago

4 0

0.02 g of calcium carbonate (CaCO₃)

Explanation:

To solve the question we use the Avogadro's number using the following reasoning:

if 1 moles of calcium carbonate contains 6.022 × 10²³ units of calcium carbonate

then X moles of calcium carbonate contains 12.044 × 10²³ units of calcium carbonate

X = (1 × 12.044 × 10²³) / 6.022 × 10²³ = 2 moles of calcium carbonate

mass = number of moles × molecular weight

mass of calcium carbonate (CaCO₃) = 2 / 100 = 0.02 g

Learn more about:

Avogadro's number

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krok68 [10]

Answer:

The organism died 35900 years ago

Explanation:

Half life C-14 is 5730 years.
For decay of radioactive nuclides: $\frac{N}{N_{0}}=(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$
$\frac{N}{N_{0}}$ is the fraction of radioactive nuclide remain present after t time and $t_{\frac{1}{2}}$ is the half-life of radioactive nuclide
Here $\frac{N}{N_{0}}$ is 0.013 and $t_{\frac{1}{2}}$ is 5730 yeras

Plug-in all the values in the above equation:

$0.013=(\frac{1}{2})^{\frac{t}{5730years}}$

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3 years ago

A sample of an unknown substance has a mass of 42g and a volume of

hjlf

Answer: 1.5 g/cm^3

Explanation:

Density is mass over volume

42g / 28 cm^3 = 1.5 g/cubic cm

hope this helps

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3 years ago

Read 2 more answers

Angela does 50 Joules of work to lift her 2.5 kg grocery bag. How far (distance) did she lift the grocery bags?

levacccp [35]

If:

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3 years ago

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Mashcka [7]

Answer:

The voltage is $E_{cell} = 0.944 \ V$

Explanation:

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substituting values

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