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3 years ago

How many moles are there in 2.00 x 10^19 molecules of CCl4

Chemistry

1 answer:

HACTEHA [7]3 years ago

7 0

Answer: 0.0000332mol

Explanation: 1mole of CCl4 contains 6.02x10^23 molecules.

Therefore, X mol of CCl4 will contain 2 x 10^19 molecules i.e

Xmol of CCl4 = 2 x 10^19/ 6.02x10^23 = 0.0000332mol

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Which organism is a tertiary consumer? clownfish brittle star killer whale octopus

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Answer:

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Explanation:

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3 years ago

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To make a 2.0-molar solution, how many moles of solute must be dissolved in 0.50 liters of solution?

attashe74 [19]

Moles = Molarity x Volume

Moles = 2.0 x 0.50

= 1.0 mole

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2 years ago

Stomach acid is approximately 0.10 M HCl. How many mL of stomach acid can be neutralized by one regular antacid tablet that cont

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Answer:

100 mL

Explanation:

The reaction that takes place is:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

First we convert 500 mg of CaCO₃ into mmoles, using its molar mass:

500 mg ÷ 100 mg/mmol = 5 mmol CaCO₃

Then we convert 5 mmoles of CaCO₃ into HCl mmoles, using the stoichiometric coefficients of the balanced reaction:

5 mmol CaCO₃ * $\frac{2mmolHCl}{1mmolCaCO_3}$ = 10 mmol HCl

Finally we calculate the volume of a 0.10 M HCl solution (such as stomach acid) that would contain 10 mmoles:

10 mmol / 0.10 M = 100 mL

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2 years ago

When does the first step of digestion usually occur?

SVEN [57.7K]

Answer:

when the food is chewed

Explanation:

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2 years ago

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Please help me solve this!

yulyashka [42]

Answer : The image is attached below.

Explanation :

For $O_3$ :

Molar mass, M = 48 g/mol

Mass, m = 24 g

Moles, n = $\frac{m}{M}=\frac{24g}{48g/mol}=0.5mol$

Number of particles, N = $n\times 6.022\times 10^{23}=0.5\times 6.022\times 10^{23}=3.0\times 10^{23}$

For $NH_3$ :

Molar mass, M = 17 g/mol

Mass, m = 170 g

Moles, n = $\frac{m}{M}=\frac{170g}{17g/mol}=10mol$

Number of particles, N = $n\times 6.022\times 10^{23}=10\times 6.022\times 10^{23}=6.0\times 10^{24}$

For $F_2$ :

Molar mass, M = 38 g/mol

Mass, m = 38 g

Moles, n = $\frac{m}{M}=\frac{38g}{38g/mol}=1mol$

Number of particles, N = $n\times 6.022\times 10^{23}=1\times 6.022\times 10^{23}=6.0\times 10^{23}$

For $CO_2$ :

Molar mass, M = 44 g/mol

Moles, n = 0.10 mol

Mass, m = $n\times M=0.10mol\times 44g/mol=4.4g$

Number of particles, N = $n\times 6.022\times 10^{23}=0.10\times 6.022\times 10^{23}=6.0\times 10^{22}$

For $NO_2$ :

Molar mass, M = 46 g/mol

Moles, n = 0.20 mol

Mass, m = $n\times M=0.20mol\times 46g/mol=9.2g$

Number of particles, N = $n\times 6.022\times 10^{23}=0.20\times 6.022\times 10^{23}=1.2\times 10^{23}$

For $Ne$ :

Molar mass, M = 20 g/mol

Number of particles = $1.5\times 10^{23}$

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{1.5\times 10^{23}}{6.022\times 10^{23}}=0.25mol$

Mass, m = $n\times M=0.25mol\times 20g/mol=5g$

For $N_2O$ :

Molar mass, M = 44 g/mol

Number of particles = $1.2\times 10^{24}$

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{1.2\times 10^{24}}{6.022\times 10^{23}}=1.9mol$

Mass, m = $n\times M=1.9mol\times 44g/mol=83.6g$

For unknown substance:

Number of particles = $3.0\times 10^{23}$

Mass, m = 8.5 g

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{3.0\times 10^{23}}{6.022\times 10^{23}}=0.50mol$

Molar mass, M = $\frac{m}{n}=\frac{8.5g}{0.50mol}=17g/mol$

The substance is $NH_3$ .

3 0

2 years ago