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3 years ago

Which of the following natural resources cant be reused

1 answer:

6 0

Hi there! Air and sunlight can definitely be reused. Those are abundant and renewable resources. Therefore, A and D are eliminated. There is a limited amount of water, however, it's impossible to run out of it to the point that there's no more on Earth. C is out. The only answer choice that makes sense is coal, because it's a nonrenewable resource, and it takes millions of years to make more of. It's a fossil fuel, so once we use them up, we can't get anymore during our lives. The answer is B: coal.

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Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. K2SO4 NH4I CoCl3

katrin [286]

The substance that releases the greatest amount of ions will have the greatest attractive forces within its solution, resulting in a reduced freezing point.
K₂SO₄ yields 3 ions
NH₄I yields 2 ions
CoCl₃ yields 4 ions

Freezing points:
CoCl₃ < K₂SO₄ < NH₄I

8 0

3 years ago

Se tiene una solución acuosa 2M de carbonato de potasio. Expresar su concentración en %p/v y Normalidad.

zepelin [54]

Answer:

Normalidad = 4N

%p/V = 27.6%

Explanation:

La solución 2M de carbonato de potasio contiene 2moles de carbonato por litro de solución. La normalidad son los equivalente de carbonato de potasio (2eq/mol) por litro de solución:

2moles * (2eq/mol) = 4eq / 1L = 4N

El porcentaje peso volumen es el peso de carbonato en gramos dividido en el volumen en mL por 100:

%p/V:

Masa K2CO3 -Masa molar: 138.205g/mol-

2moles * (138.205g/mol) = 276g K2CO3

Volumen:

1L * (1000mL/1L) = 1000mL

%p/V:

276g K2CO3 / 1000mL * 100

6 0

3 years ago

Which formula can be used to calculate the molar mass of hydrogen peroxide (H2O2)? (5 points)

ratelena [41]

answer: option d (2×molar mass of H +2×molar mass of O

 

7 0

3 years ago

Some people must eat a low-sodium diet with no more than 2,000 mg of sodium per day. By eating 1 cracker, 1 pretzel, and 1 cooki

Leni [432]

Answer:

The amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

Explanation:

Let's assume amount of sodium is x mg per cracker, y mg per pretzel and z mg per cookie.

So, the following three equations can be written as per given information:

x+y+z = 149 ........(1)

8y+8z = 936 ........(2)

6x+7y = 535 .........(3)

From equation- (2), y+z = $\frac{936}{8}$ = 117

By substituting the value of (y+z) in equation- (1) we get,

x = 149-(y+z) = 149-117 = 32

By substituting the value of x into equation- (3) we get,

y = $\frac{535-(6\times 32)}{7}$ = 49

By substituting the value of y into equation- (2) we get,

z = (117-49) = 68

So, the amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

6 0

3 years ago

Part A

Elden [556K]

Answer:

ΔG° = -5.4 kJ/mol

ΔG = 873.2 J/mol = 0.873 kJ /mol

Explanation:

Step 1: Data given

ΔG (NO2) = 51.84 kJ/mol

ΔG (N2O4) = 98.28 kJ/mol

Step 2:

ΔG = ΔG° + RT ln Q

⇒with Q = the reaction quatient

⇒with T = the temperature = 298 K

⇒with R = 8.314 J / mol*K

⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2 )

⇒ ΔG° = 98.28 kJ/mol - 2* 51.84 kJ/mol

⇒ ΔG° = -5.4 kJ/mol

Part B

ΔG = ΔG° =RT ln Q

⇒with G° = -5.4 kj/mol = -5400 j/mol

⇒ with R = 8.314 J/K*mol

⇒with T = 298 K

⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577

ΔG = -5400 + 8.314 * 298 * ln(12.577)

ΔG = -5400 + 8.314 * 298 * 2.532

ΔG = 873.2 J/mol = 0.873 kJ/mol

3 0

3 years ago