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NISA [10]
4 years ago
9

1. Name the two types of low-level clouds.

Chemistry
1 answer:
Mice21 [21]4 years ago
4 0
Nimbostratus and Stratocumulus are two types of low-level clouds!
You might be interested in
Examples of symmetrical objects
butalik [34]

Answer:

Explanation:

Reflection of trees in clear water and reflection of mountains in a lake. Wings of most butterflies are identical on the left and right sides. Some human faces are the same on the left and right side. People can also have a symmetrical mustache.

8 0
3 years ago
Read 2 more answers
Benzene is 92.3% carbon and 7.7% hydrogen. In anexperiment,
Marysya12 [62]

Answer:

The molecular formula of benzene = C_{6}H_{6}

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

% of C = 92.3

Molar mass of C = 12.0107 g/mol

% moles of C = \frac{92.3}{12.0107} = 7.6848

% of H = 7.7

Molar mass of H = 1.00784 g/mol

% moles of H = \frac{7.7}{1.00784} = 7.6401

Taking the simplest ratio for C and H as:

7.6848 : 7.6401

= 1 : 1

The empirical formula is = CH

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12+ 1 = 13 g/mol

Molar mass = 78.0 g/mol

So,  

Molecular mass = n × Empirical mass

78.0 = n × 13

⇒ n ≅ 6

<u>The molecular formula of benzene = C_{6}H_{6}</u>

7 0
4 years ago
How much energy is required to heat 87.1 g acetone (molar mass=58.08 g/mol) from a solid at -154.0°C to a liquid at -42.0°C? The
WARRIOR [948]

Answer:

The answer to the question above is

The energy required to heat 87.1 g acetone from a solid at -154.0°C to a liquid at -42.0°C = 29.36 kJ

Explanation:

The given variables are

ΔHfus = 7.27 kJ/mol

Cliq = 2.16 J/g°C

Cgas = 1.29 J/g°C

Csol = 1.65 J/g°C

Tmelting = -95.0°C.

Initial temperature = -154.0°C

Final temperature = -42.0°C?

Mass of acetone = 87.1 g

Molar mass of acetone = 58.08 g/mol

Solution

Heat required to raise the temperature of solid acetone from -154 °C to -95 °C or 59 °C is given by

H = mCsolT = 87.1 g* 1.65 J/g°C* 59 °C = 8479.185 J

Heat required to melt the acetone at -95 °C = ΔHfus*number of moles =

But number of moles = mass÷(molar mass) = 87.1÷58.08 = 1.5

Heat required to melt the acetone at -95 °C =1.5 moles*7.27 kJ/mol = 10.905 kJ

The heat required to raise the temperature to -42 degrees is

H = m*Cliq*T = 87.1 g* 2.16 J/g°C * 53 °C = 9971.21 J

Total heat = 9971.21 J + 10.905 kJ + 8479.185 J = 29355.393 J = 29.36 kJ

The energy required to heat 87.1 g acetone from a solid at -154.0°C to a liquid at -42.0°C is 29.36 kJ

4 0
4 years ago
Please explain, I don't understand.
BartSMP [9]

Answer:

7. A) I, II ; 8. D) 2.34e9 kJ

Step-by-step explanation:

7. Combustion of ethanol

I. The negative sign for ΔH shows that the reaction is exothermic.

II. The enthalpy change would be different if gaseous water were produced.

That's because it takes energy to convert liquid water to gaseous water, and this energy is included in the value of ΔH.

III. The reaction is a redox reaction, because

  • Oxygen is reacting with a compound
  • The oxidation number of C increases
  • The oxidation number of O decreases.

IV. The products of the reaction occupy a smaller volume than the reactants, because 3 mol of gaseous reactant are forming 2 mol of gaseous product.

Therefore, only I and II are correct.

7. Hindenburg

Data:

  V = 2.00 × 10⁸ L

  p = 1.00 atm

  T = 25.1 °C

ΔH = -286 kJ·mol⁻¹

Calculations:

(a) Convert temperature to kelvins

T = (25.1 + 273.15) K = 298.25 K

(b) Moles of hydrogen

Use the <em>Ideal Gas Law</em>:

pV = nRT

n = (pV)/(RT)

n = (1.00 × 2.00 × 10⁸)/(0.082 06 × 298.25) = 8.172 × 10⁶ mol

(c) Heat evolved

q = nΔH = 8.172 × 10⁶ × (-286) = -2.34 × 10⁹ kJ

The hydrogen in the Hindenburg released 2.34e9 kJ .

4 0
4 years ago
Determine the pH of a 0.015 M HNO3 solution.
miskamm [114]

Answer:

0.015 M HNO3 = 0.015 M H3O+ pH = -log(0.015) = 1.82 [OH -] = 1.0 x 10 -14/0.015 M = 6.7 x 10 13 Check: HNO3 is a strong acid, so we expect the pH to be low and the [OH ] to be small

8 0
3 years ago
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