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3 years ago

Please Help Thank You

Chemistry

2 answers:

Karolina [17]3 years ago

7 0

W = F * d
F = 600; d = 0
W = 600 * 0 = 0J

natka813 [3]3 years ago

6 0

The answer is 600N because if its 600 and 0 its still going to be 600

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Calculate the number of pounds of CO2 released into the atmosphere when a 17.0-gallon tank of gasoline is burned in an automobil

rosijanka [135]

I have no clue at all hope someone can help u

8 0

3 years ago

Please answer both questions Thankyou

erma4kov [3.2K]

Answer:

The limiting reactant is oxygen gas and the reaction would produce 3.932 x 10 ^ 7 moles of water

Explanation:

Step 1: Convert everything into moles

nH2(l) = 1.06 x 10^8 g / 2.016 g/mol = 5.26 x 10^7 mols

nO2(l) = 6.29 x 10^8 g / 32.00 g/ mol = 1.966 x 10^7 mols

Step 2: Find the limiting reagent

The limiting reagent would be oxygen gas from

the balanced equation because we have less moles of oxygen gas needed to fully combust with the hydrant gas

Step 3: Stoichiometry time

The mole ratio from oxygen gas to water is 1:2

This means that for every mole of oxygen gas two moles of water is produced

We need to multiply the moles of oxygen gas by two to find out how many moles of water has been produced

nH2O = nO2 x 2

nH2O = 1.966x10^7 x 2

nH2O = 3.932x10^7

Step 4: Therefore statement

Therefore the limiting reactant is oxygen gas and the reaction would produce 3.932 x 10 ^ 7 moles of water

7 0

3 years ago

Steam reforming of methane ( CH4 ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the sta

Hatshy [7]

Answer: The pressure equilibrium constant for the reaction is 1473.8

Explanation:

We are given:

Initial partial pressure of methane gas = 2.4 atm

Initial partial pressure of water vapor = 3.9 atm

Equilibrium partial pressure of hydrogen gas = 6.5 atm

The chemical equation for the reaction of methane gas and water vapor follows:

$CH_4+H_2O\rightleftharpoons CO+3H_2$

Initial: 2.4 3.9

At eqllm: 2.4-x 3.9-x x 3x

Evaluating the value of 'x':

$\Rightarrow 3x=6.5\\\\x=2.167$

So, equilibrium partial pressure of methane gas = (2.4 - x) = [2.4 - 2.167] = 0.233 atm

Equilibrium partial pressure of water vapor = (3.9 - x) = [3.9 - 2.167] = 1.733 atm

Equilibrium partial pressure of carbon monoxide gas = x = 2.167 atm

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{CO}\times (p_{H_2})^3}{p_{CH_4}\times p_{H_2O}}$

Putting values in above equation, we get:

$K_p=\frac{2.167\times (6.5)^3}{0.233\times 1.733}\\\\K_p=1473.8$

Hence, the pressure equilibrium constant for the reaction is 1473.8

5 0

3 years ago

Provide three specific examples of how this cloning is used.

lara [203]

Answer:

Cloning is used to duplicate different organisms

5 0

3 years ago

You perform a single extraction with 100 mL of chloroform to remove compound A from 100 mL of an aqueous solution. The partition

Mice21 [21]

Answer:

Percentage of mass of solute in aqueous phase is 28.57%.

Explanation:

$(q)_{aq}$ - Fraction of solute in aqueous phase.

$(q)_{aq}=\frac{V_{aqueous}}{DV_{organic}+V_{aqueous})}$

$V_{aqueous}$ = Used volume of aqueous solution

$V_{organic}$ = Used volume of organic solution

D = partition coefficient.

$(q)_{aq}=\frac{100ml}{(2.5\times100)+100\,ml}=0.2857$

Mass of solute in aqueous phase = $0.2857\times 5g=1.428g$

$Mass\,percentage = \frac{1.428}{5.0}\times 100=28.57$

Therefore, Percentage of mass of solute in aqueous phase is 28.57%.

5 0

3 years ago