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2 years ago

Yo yuh guys know this

Chemistry

1 answer:

erastovalidia [21]2 years ago

8 0

Answer:80KM is distance. 30KM north is displacement.

Explanation:

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Write a question about how changing temperature affects a gas.

UNO [17]

Which property of gas affects the gas by change in that property ?

7 0

2 years ago

Read 2 more answers

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

2 years ago

What is the mass percent of barium in barium cyanide?

GaryK [48]

Answer:

72.521%

Explanation:

7 0

2 years ago

A substance is analyzed and found to contain 85.7% carbon and 14.3% hydrogen by weight. A gaseous sample of the substance is fou

atroni [7]

Answer:

The empirical formula of the compound is CH2

Explanation:

Step 1: Data given

A substance contains 85.7 % carbon and 14.3 % hydrogen.

The substance has a density of 1.87 g/L

1 mol occupies 22.4 L

Molar mass of carbon = 12 g/mol

Molar mass of hydrogen = 1.01 g/mol

Step 2: Calculate molar mass of the substance

Since 1 mol occupies 22.4 L;

1 mol of this substance = 1.87g/L *22.4 = 41.888 grams

This means the molar mass of the substance is 41.888 g/mol

Step 3: Calculate mass of carbon:

85.8 % is carbon

this means 41.888 * 0.858 = 35.94 grams

Step 4: Calculate moles of carbon

moles C = mass C/ Molar mass C

Moles C = 35.94 grams / 12 g/mol

Moles C = 2.995 moles

Step 5: Calculate mass of hydrogen:

14.3 % is hydrogen

this means 41.888 * 0.143 = 5.99 grams

Step 6 :Calculate moles of hydrogen

Moles H = 5.99 grams / 1.01 g/mol

Moles H = 5.93 moles

Step 7: Calculate mol ratio

Ratio C:H = 1:2

The empirical formule = CH2

Step 8: calculate molar formule

Molar mass of empirical formule = 14.02 g/mol

n = Molar mass of substance / molar mass of empirical formule

n = 41.888 / 14.02 = 3

This means we have to multiply the empirical formula by 3

3*(CH2) = C3H6

C3H6 can be propene or cyclopropane

5 0

2 years ago

Why might it be important to follow the chain of custody when handling evidence?

Ierofanga [76]

Answer:

To keep a record of who has the evidence and when

Explanation:

Don't want anyone tampering with it, hope this helps!!!

5 0

2 years ago