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butalik [34]
3 years ago
12

I need helpppp! i don’t understand this

Chemistry
1 answer:
Marta_Voda [28]3 years ago
4 0

Answer: a. 410 K

b. 296 K

c. 394 K

d. -180 °C

e. 224°C

f. -3 °C

Explanation:

For °C to K : use °C + 273 = K

For K to °C : use °C = K - 273

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Calculate the percent by mass of 4.35g of Na I dissolved in 105g of water​
Ludmilka [50]

<u>We are given:</u>

Mass of Na added = 4.35 grams

Mass of water = 105 grams

<u>Mass Percent of Na:</u>

Total mass of the solution = mass of solute + mass of solvent

Total mass of the solution = 4.35 + 105 = 109.35 grams

Mass percent of solute = (mass of solute / mass of solution) * 100

Mass percent of Solute = (4.35 / 109.35) * 100

Mass percent = 3.978 %

4 0
3 years ago
Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t
myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol

Moles of methane gas = n_1=\frac{8.00 g}{16 g/mol}=0.5 mol

Moles of ethane gas  =n_2=\frac{18.0 g}{30 g/mol}=0.6 mol

Moles of propane gas = n_3

n=n_1+n_2+n_3

2.22=0.5 mol +0.6 mol+ n_3

n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol

Mole fraction of methane =\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}

\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252

Similarly, mole fraction of ethane and propane :

\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703

\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045

Partial pressure of each gas can be calculated by the help of Dalton's' law:

p_i=P\times \chi_1

Partial pressure of methane gas:

p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm

Partial pressure of ethane gas:

p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm

Partial pressure of propane gas:

p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = n_1=\frac{37.8 g g}{28g/mol}=1.35 mol

Moles of oxygen gas  =n_2=\frac{62.2 g}{32 g/mol}=1.94 mol

Mole fraction of nitrogen=\chi_1=\frac{n_1}{n_1+n_2}

\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103

Similarly, mole fraction of oxygen

\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897

Partial pressure of each gas can be calculated by the help of Dalton's' law:

p_i=P\times \chi_1

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg

Partial pressure of oxygen gas:

p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg

3 0
3 years ago
How is the frequency of an EM wave related to its energy
natulia [17]

Answer:

Higher the frequency, the higher the energy

Explanation:

Please help me by marking me brainliest. I'm really close :)

6 0
3 years ago
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Identify the substance that conducts electricity. question 19 options: 1) nacl dissolved in water 2) solid nacl 3) solid sugar 4
Delicious77 [7]

The substance that conducts electricity is NaCl dissolved in water.

So, option A is correct one.

When the sodium chloride dissolve in water , the sodium atoms and chlorine atoms separates under the presence of water molecules and exist as sodium cation and chloride anion . Now , they are free to move around in the water as positively and negatively charged ions . This separation of charge allows the solution to conduct electricity.

The solid NaCl and solid sugar does not conduct electricity because it is not dissolve in water . Similarly, sugar dissolved in water does not conduct electricity .

to learn more about conduct electricity.

brainly.com/question/1458220

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4 0
2 years ago
What do plants do that requires Energy?
Kruka [31]

Answer:

Photosynthesis

Explanation:

Photosynthesis uses energy from light to convert water and carbon dioxide molecules into glucose (sugar molecule) and oxygen. The oxygen is released, or “exhaled”, from leaves while the energy contained within glucose molecules is used throughout the plant for growth, flower formation, and fruit development.

7 0
3 years ago
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