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3 years ago

Write the name for the following molecular compounds. Remember to use the correct prefix for each compound.

Chemistry

1 answer:

Westkost [7]3 years ago

3 0

Answer:

Hey there!

CS2) Carbon Disulfide.

PBr3) Phosphorus Tribromide

NO) Nitric Oxide

CF4) Carbon Tetrafluoride

P2O5) Phosphorus Pentoxide

Let me know if this helps :)

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Explain electronegativity and how it relates to both molecular and ionic bonding.

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Answer:

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3 years ago

Drinking water may contain a low concentration of lead ion (Pb2 ) due to corrosion of old lead pipes. The EPA has determined tha

Lynna [10]

Answer:

3.861x10⁻⁹ mol Pb⁺²

Explanation:

We can define ppm as mg of Pb²⁺ per liter of water.

We calculate the mass of lead ion in 100 mL of water:

100.0 mL ⇒ 100.0 / 1000 = 0.100 L

0.100 L * 0.0080 ppm = 8x10⁻⁴ mg Pb⁺²

Now we convert mass of lead to moles, using its molar mass:

8x10⁻⁴ mg ⇒ 8x10⁻⁴ / 1000 = 8x10⁻⁷ g

8x10⁻⁷ g Pb²⁺ ÷ 207.2 g/mol = 3.861x10⁻⁹ mol Pb⁺²

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3 years ago

Elements join together to form_____.

sineoko [7]

Answer:

compounds

Explanation:

6 0

3 years ago

Need help with the question

Sloan [31]

Answer:

for a i think it is choice 2

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Explanation:

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3 years ago

Acetylene gas, C2H2, can be produced by the reaction of calcium carbide and water. CaC2(s) + 2H2O(l) --> C2H2(g) + Ca(OH)2(aq

nekit [7.7K]

Answer:

1.0 L

Explanation:

Given that:-

Mass of $CaC_2$ = $2.54\ g$

Molar mass of $CaC_2$ = 64.099 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{2.54\ g}{64.099\ g/mol}$

$Moles_{CaC_2}= 0.0396\ mol$

According to the given reaction:-

$CaC_2_{(s)} + 2H_2O_{(l)}\rightarrow C_2H_2_{(g)} + Ca(OH)_2_{(aq)}$

1 mole of $CaC_2$ on reaction forms 1 mole of $C_2H_2$

0.0396 mole of $CaC_2$ on reaction forms 0.0396 mole of $C_2H_2$

Moles of $C_2H_2$ = 0.0396 moles

Considering ideal gas equation as:-

$PV=nRT$

where,

P = pressure of the gas = 742 mmHg

V = Volume of the gas = ?

T = Temperature of the gas = $26^oC=[26+273]K=299K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles = 0.0396 moles

Putting values in above equation, we get:

$742mmHg\times V=0.0396 mole\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 299K\\\\V=\frac{0.0396\times 62.3637\times 299}{742}\ L=1.0\ L$

1.0 L of acetylene can be produced from 2.54 g $CaC_2$ .

4 0

3 years ago