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9966 [12]
3 years ago
14

Which statement is correct about the water cycle?

Chemistry
1 answer:
OleMash [197]3 years ago
7 0

Answer:

precipitation is when water flows into the ground

Explanation:

.................

You might be interested in
Which of the following sets of empirícal formula, molar mass, and molecular formula is correct?
skad [1K]

<em>Answer:</em>

  • The option C is correct.
  • CH4N, 90g, C3H12N3.

<em>Explanation:</em>

                    <em>Option C:</em>

  • The molecular formula is C3H12N3.
  • If we take ratio 3:12:3 that will be equal to 1:4:1 so its empirical formula will be CH4N.
  • The molar mass of molecular formula = (12×3) + (1×12) +( 14×3) = 98 g

                <em>Option A :</em>

  • In option A , molecular and empirical formula are both correct but molar mass is not valid. It should be 169 acc. to molecular formula.

              <em>Option B:</em>

  • In option B, molecular formual is not valid.
  • It should be as C6H16O2
4 0
3 years ago
Two substances A and B, initially at different temperatures, are thermally isolated from their surroundings and allowed to come
Elden [556K]

Answer:

B

Explanation:

For solving this we need a heat balance

Q_{a} = Q_{b}\\m_{a}*C_{a}*\Delta T_{a} = m_{b}*C_{b}*\Delta T_{b}

By changing the corresponding relations, we have

m_{a}*C_{a}*\Delta T_{a} = \frac{1}{2}m_{a}*4C_{a}*\Delta T_{b} \\\\\\

By cancelling similar factor, we obtain

\Delta T_{a} = 2 \Delta T_{b}\\\frac{\Delta T_{a}}{\Delta T_{b}} = 2\\

Which means that the change of temperature in A is twice the change of B

3 0
3 years ago
(chem) which is more concentrated: 45.0 grams of HCOOH dissolved in 189 mL of water or 1.5 moles of CH↓2COOH dissolved in twice
Liula [17]

Answer:

CH3COOH would be more concentrated

Explanation:

The higher the concentration value, the more concentrated it is.

The relationship between concentration, moles and volume is given by the equation;

Concentration = No of moles / Volume

5.0 grams of HCOOH dissolved in 189 mL of water

Number of moles = Mass / Molar mass = 5 / 46.03 = 0.1086 mol

Concentration = 0.1086 / 0.189 = 0.5746 mol/L

1.5 moles of CH3COOH dissolved in twice as much water

Volume = 2 * 189 = 378 ml = 0.378 L

Concentration = 1.5 / 0.378 = 3.9683 mol/L

Comparing both concentration values;

CH3COOH would be more concentrated

6 0
3 years ago
When the concentration of SO2(g) is increased to 1.48 M, the ratio of products to reactants is 1.4. The equilibrium constant for
weeeeeb [17]

Answer:

Towards the products

Explanation:

3 0
3 years ago
Compute the values of the diffusion coefficients for the interdiffusion of carbon in both α-iron (BCC) and γ-iron (FCC) at 900°C
bogdanovich [222]

Answer:

α-iron (BCC) has faster diffusion rate because of lower values in activation energy and pre-exponential value.

Explanation:

Taking each parameters or data at a time, we can determine the values/a constant for each parameters in the diffusion coefficient equation.

For α-iron (BCC), the diffusion coefficient = pre-exponential value,Ao × e^( -Activation energy,AE)/gas constant,R × Temperature.

Converting the given Temperature, that is 900°C to Kelvin which is equals to 1173.15K.

For α-iron (BCC), the pre-exponential value, Ao = 1.1 × 10^-6, and the activation energy, AE = 87400.

Thus, we have that the diffusion coefficient = 1.1 × 10^-6 × e(-87400)/1173.15 × 8.31.

Diffusion coefficient for α-iron (BCC) = 1.41 × 10^-10 m^2/s.

Also, For the γ-iron (FCC), the pre-exponential value, Ao = 2.3 × 10^-5 and the activation energy, AE = 148,00.

From these values we can see that both the exponential value, Ao and the activation energy for γ-iron (FCC) are higher than that of α-iron (BCC).

Thus, the diffusion coefficient for the γ-iron (FCC) = 2.3 × 10^-5 × e ^-(14800)/8.31 × 1173.15.

Then, the diffusion coefficient for the γ-iron (FCC) = 5.87 × 10^-12 m2/s.

Therefore, there will be faster diffusion in α-iron (BCC) because of lower activation energy and vice versa.

6 0
3 years ago
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