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3 years ago

I need help on 14-17

Chemistry

1 answer:

Zarrin [17]3 years ago

3 0

14 B, 15 A, 16 B, 17 B.

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3 years ago

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CF4 + Br2 CBr4 + F2

luda_lava [24]

Answer:

Answers are in the explanation

Explanation:

Based on the reaction:

CF₄ + 2Br₂ → CBr₄ + 2F₂

The mole ratio of CF₄ is:

CF₄:Br₂ = 1:2

CF₄:CBr₄ = 1:1

CF₄:F₂ = 1:2

<em>Moles F2:</em>

Molar mass CF₄: 88.0g/mol

57.0g * (1mol / 88.0g) = 0.6477 moles CF₄ * (2mol F₂ / 1mol CBr₄) =

<h3>1.30 moles F₂</h3><h3 />

Molar mass CBr₄: 331.63g/mol

250.0g * (1mol / 331.63g) = 0.7539 moles CBr₄ * (2mol Br₂ / 1mol CF₄) =

1.51 moles Br₂ * (159.808g / mol) =

<em>Moles F2:</em>

4.8 moles CF₄ * (2mol F₂ / 1mol CF₄) =

<h3>9.6 moles F₂</h3><h3 />

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3 years ago

Somebody help me asap please giving brainliest

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3 years ago

When the volume of a gas is

Montano1993 [528]

Answer:

$\boxed {\boxed {\sf 232.9 \textdegree C}}$

Explanation:

This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

$\frac {V_1}{T_1}= \frac{V_2}{T_2}$

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

$\frac{250 \ mL}{137 \textdegree C}= \frac{V_2}{T_2}$

The volume of the gas is increased to 425 milliliters, but the temperature is unknown.

$\frac{250 \ mL}{137 \textdegree C}= \frac{425 \ mL}{T_2}$

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

$250 \ mL * T_2 = 137 \textdegree C * 425 \ mL$