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2 years ago

Heat, like work, is an energy transfer measured in a. watts. b. degrees. c. joules. d. Kelvins.

Chemistry

2 answers:

Vesna [10]2 years ago

8 0

Answer:

Joules

Explanation:

Hope I could help

Vesna [10]2 years ago

6 0

The answer is A watts

Glad I could help

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Identify the intermolecular forces present in each of these substances nh3 hcl co co2

nasty-shy [4]

Intermolecular forces are forces of interaction that are operative between two different molecules. They are of follow types
1) Dipole- dipole interaction
2) Hydrogen bonds
3) vander Waal's forces

Depending upon the polarity and constitution molecule above forces are operable.

for instance, in case of ammonia, Hydrogen bonds exist because hydrogen atom is attached to electronegative element i.e. N

HCl and CO are polar molecules, so dipole-dipole interaction is operative in these molecules.

Finally in case of CO2, vander Waal's forces of interaction is operable because it is a non-polar molecule.

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2 years ago

What volume of carbon dioxide, at 1 atm pressure and 112°C, will be produced when 80.0 grams of methane is burned?

Vadim26 [7]

Answer:

158 L.

Explanation:

What is given?

Pressure (P) = 1 atm.

Temperature (T) = 112 °C + 273 = 385 K.

Mass of methane CH4 (g) = 80.0 g.

Molar mass of methane CH4 = 16 g/mol.

R constant = 0.0821 L*atm/mol*K.

What do we need? Volume (V).

Step-by-step solution:

To solve this problem, we have to use ideal gas law: the ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. The formula is:

$PV=nRT.$

Where P is pressure, V is volume, n is the number of moles, R is the constant and T is temperature.

So, let's find the number of moles that are in 80.0 g of methane using its molar mass. This conversion is:

$80.0g\text{ CH}_4\cdot\frac{1\text{ mol CH}_4}{16\text{ g CH}_4}=5\text{ moles CH}_4.$

So, in this case, n=5.

Now, let's solve for 'V' and replace the given values in the ideal gas law equation:

$V=\frac{nRT}{P}=\frac{5\text{ moles }\cdot0.0821\frac{L\cdot atm}{mol\cdot K}\cdot385K}{1\text{ atm}}=158.04\text{ L}\approx158\text{ L.}$

The volume would be 158 L.

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7 months ago