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3 years ago

Complete the passage.

Chemistry

2 answers:

EastWind [94]3 years ago

4 0

Answer:

Sound waves and some earthquake waves are longitudinal waves.Ocean, light, and other earthquake waves are transverse waves

Explanation:

Natalka [10]3 years ago

3 0

Answer:

Longitudinal and Transverse

Explanation:

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What mass of mgf2 is contained in 80.85 g of a 22.4% by mass solution of mgf2 in water?

Aleks [24]

Formula: % by mass = (mass of solute / mass of solution] *100

Data:

mass of solution = 80.85 g

% by mass = 22.4%

Unknown = mass of solute

Solution

% by mass = (mass of solute / mass of solution] *100 = >

mass of solute = % by mass * mass of solution / 100

mass of solute = 22.4 * 80.85 / 100 = 18.11 g

Answer: 18.11 g

4 0

3 years ago

Read 2 more answers

Sulfuric acid is essential to dozens of important industries from steelmaking to plastics and pharmaceuticals. More sulfuric aci

Pavlova-9 [17]

Given:

K = 0.71 = Kp

The reaction of sulphur with oxygen is

S(s) + O2(g) ---> SO2(g)

initial Pressure 6.90 0

Change -x +x

Equilibrium 6.90-x x

Kp = pSO2 / pO2 = 0.71 = x / (6.90-x)

4.899 - 0.71x = x

4.899 = 1.71x

x = 2.86 atm = pressure of SO2 formed

temperature = 950 C = 950 + 273.15 K = 1223.15 K

Volume = 50 L

Let us calculate moles of SO2 formed using ideal gas equation as

PV = nRT

R = gas constant = 0.0821 L atm / mol K

putting other values

n = PV / RT = 2.86 X 50 / 1223.15 X 0.0821 = 1.42 moles

Moles of Sulphur required = 1.42 moles

Mass of sulphur required or consumed = moles X atomic mass of sulphur

mass of S = 1.42 X 32 = 45.57 grams or 0.04557 Kg of sulphur

6 0

3 years ago

At 338 mm hg and 72 c a sample of carbon monoxide gas occupies a volune of 0.225 L the gas transferred to a 1.50 L flask and the

Elis [28]

Answer:

P₂ = 0.09 atm

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 0.225 L

Initial pressure = 338 mmHg (338/760 =0.445 atm)

Initial temperature = 72 °C (72 +273 = 345 K)

Final temperature = -15°C (-15+273 = 258 K)

Final volume = 1.50 L

Final pressure = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

P₂ = P₁V₁ T₂/ T₁ V₂

P₂ = 0.445 atm × 0.225 L × 258 K / 345 K × 1.50 L

P₂ = 25.83 atm .L. K / 293 K . L

P₂ = 0.09 atm

7 0

4 years ago

What is the relationship between the amount of carbon dioxide and the earths average temperature?

Anit [1.1K]

Answer:

carbon dioxide concentration goes down, temperature goes down. Carbon dioxide goes up temp goes up, carbon dioxide is directly related to temperature by insulating it in the earths atmosphere and if there's less of it then the temp would go down.

Explanation:

6 0

2 years ago

g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc

Gala2k [10]

$\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}$

$\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}$

$\text{and in the liquid form it is easily transported. An industrial chemist studying this}$

$\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }$

$\text{to be} \ \mathbf{2.6\ mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}$

$\text{ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.}$

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

From the above reaction; the ICE table can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

I (mol/L) 0.086 0.28 0 0

C -4x -3x +2x +6x

E 0.086 - 4x 0.28 - 3x +2x +6x

At equilibrium;

The water vapor = $\dfrac{2.6 \ mol}{100 \ L} = 6x$

$x = \dfrac{2.6}{100} \times \dfrac{1}{6}$

$x = 0.00433$

$\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }$

$\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\$

Replacing the value of x, we have:

$K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}$

$K_c = \mathbf{5.5 \times 10^{-8} \ to \ 2 \ significant \ figures}$

5 0

3 years ago