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3 years ago

Which states of matter can flow?

Chemistry

2 answers:

Iteru [2.4K]3 years ago

7 0

Answer:

1) Gas and Liquid

Explanation:

hope it helps

vladimir1956 [14]3 years ago

6 0

Yah it’s the first one

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In a chemical reaction at constant temperature, the addition of a catalyst: _______________

Tom [10]

Answer:

affects the equilibrium content

4 0

3 years ago

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The volume of gas is 252 mL at 340.0 kPa pressure. What will the volume be when the pressure is reduced to 50.0 kPa? Temperature

MrRissso [65]

Answer:

Clearly volume will INCREASE almost sevenfold, as we would expect if we reduce the pressure. Marilia E.K. The final volume of this gas will be 1.7L.

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3 years ago

1. For the reaction 3A — C, the initial concentration of A was 0.2 M, and the reaction rate was

Andrew [12]

Answer:

$r=25M^{-1}s^{-1}[A]^2$

Explanation:

Hello there!

In this case, according to the given information for this chemical reaction, it is possible for us to set up the following general rate law and the ratio of the initial and the final (doubled concentration) condition:

$r=k[A]^n\\\\\frac{r_1}{r_2} =\frac{k[A]_1^n}{k[A]_2^n}$

Next, we plug in the given concentrations of A, 0.2M and 0.4 M, the rates, 1.0 M/s and 4.0 M/s and cancel out the rate constants as they are the same, in order to obtain the following:

$\frac{1.0}{4.0} =\frac{0.2^n}{0.4^n}\\\\0.25=0.5^n\\\\n=\frac{ln(0.25)}{ln(0.5)} \\\\n=2$

Which means this reaction is second-order with respect to A. Finally, we calculate the rate constant by using n, [A] and r, to obtain:

$k=\frac{r}{[A]^n} =\frac{1.0M/s}{(0.2M)^2}\\\\k=25M^{-1}s^{-1}$

Thus, the rate law turns out to be:

$r=25M^{-1}s^{-1}[A]^2$

Regards!

3 0

3 years ago

5. Given the following equation: 8 Fe S8 8 Fes, what mass of iron is needed to react with16.0 grams of sulfur? How many grams of

Snezhnost [94]

18 grams of fes is produced

5 0

4 years ago

In a laboratory, 1.55mg of an organic compound containing carbon, hydrogen, and oxygen is burned for analysis. This combustion r

Serjik [45]

Answer:

CH₃O

Step-by-step explanation:

1. Calculate the mass of each element

Mass of C = 1.45 mg CO₂ × (12.01 mg C/44.01 mg CO₂) = 0.3957 mg C

Mass of H = 0.89 mg H₂O × (2.016 mg H/18.02 mg H₂O) = 0.0996 mg H

Mass of O = Mass of compound - Mass of C - Mass of H = (1.55 – 0.3957 – 0.0996) mg = 1.055 mg

2. Calculate the moles of each element

Moles of C = 0.3957 mg C × 1mmol C/12.01 mg C = 0.03295 mol C

Moles of H = 0.0996 mg H × 1 mmol H/1.008 mg H = 0.0988 mol H

Moles of O = 1.055 mg O × 1 mmol O/ 16.00 mg O = 0.06592 mol O

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

C: 0.032 95/0.032 95 = 1

H: 0.0988/0.032 95 = 2.998

O: 0.065 92/0.032 95 = 2.001

4. Round the ratios to the nearest integer

C:H:O = 1:3:2

5. Write the empirical formula

The empirical formula is CH₃O.

4 0

4 years ago