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pshichka [43]
3 years ago
8

Someone please help me

Chemistry
1 answer:
Ymorist [56]3 years ago
7 0
The correct answer is B.14.007
You might be interested in
If a gas occupies 79.5 mL at -1.4°C, what temperature, in Kelvin, would it
Anna35 [415]

Answer:

121 K

Explanation:

Step 1: Given data

  • Initial volume (V₁): 79.5 mL
  • Initial temperature (T₁): -1.4°C
  • Final volume (V₂): 35.3 mL

Step 2: Convert "-1.4°C" to Kelvin

We will use the following expression.

K = °C + 273.15 = -1.4°C + 273.15 = 271.8 K

Step 3: Calculate the final temperature of the gas (T₂)

Assuming ideal behavior and constant pressure, we can calculate the final temperature of the gas using Charles' law.

V₁/T₁ = V₂/T₂

T₂ = V₂ × T₁/V₁

T₂ = 35.3 mL × 271.8 K/79.5 mL = 121 K

5 0
3 years ago
For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided
SVETLANKA909090 [29]

Answer:

\frac{1}{24}

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = \frac{1}{6}

So, fraction of the whole athletic field reserved for each fifth class = \frac{1}{4}(\frac{1}{6})=\frac{1}{24}

3 0
3 years ago
Two elements in the same period have the same number of _____ _____ in their electron clouds.
shutvik [7]

Answer:

outer electrons

Explanation:

3 0
3 years ago
Read 2 more answers
Explain why chloro acetic acid<br>strong and than acetic acid<br>​
Basile [38]

Chloroacetic acid is stronger than acetic acid because of the electron-withdrawing effect of chlorine. This effect is caused by the electronegativity.

7 0
3 years ago
Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e
igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em>  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

Explanation:

1) Moles of NaCl ,n_1=1 mol

Mass of water = m= 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute(NaCl)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p=17.19 Torr

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose ,n_1=1mol

Mass of water = m  = 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p'

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute ( glucose)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p'=17.19 Torr

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

3 0
3 years ago
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