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rjkz [21]
1 year ago
7

A student stores 20.0 moles of gas particles in a flexible container. The student then uses a pump to remove 10.0 moles of gas p

articles. The temperature and pressure remain constant. Which of the following is a true statement?
Chemistry
2 answers:
pav-90 [236]1 year ago
5 0

The volume of the container will decrease.

Sergeeva-Olga [200]1 year ago
3 0
The volume of the container will decrease. I believe this answer is correct because for the temperature and pressure to remain the same but there to be less particles the container must shrink to suit.
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All of the following reactions can be described as displacement reactions except:____________.
Lady_Fox [76]

Answer:

b

Explanation:

The reaction that is not a displacement reaction from all the options is C_6H_6_{(l)} + Cl_{2(g)} --> C_6H_5Cl_{(l)} + HCl_{(g)}

In a displacement reaction, a part of one of the reactants is replaced by another reactant. In single displacement reactions, one of the reactants completely displaces and replaces part of another reactant. In double displacement reaction, cations and anions in the reactants switch partners to form products.

<em>Options a, c, d, and e involves the displacement of a part of one of the reactants by another reactant while option b does not.</em>

Correct option = b.

8 0
1 year ago
In May 2016, William Trubridge broke the world record in free diving (diving underwater without the use of supplemental oxygen)
Maru [420]

Answer:

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

Explanation:

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 3.6 L  

V₂ = ?

P₁ = 1.0 atm

P₂ = 13.3 atm (From correct source)

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{1.0\ atm}\times {3.6\ L}={13.3\ atm}\times {V_2}

{V_2}=\frac{{1.0}\times {3.6}}{13.3}\ L

{V_2}=0.27\ L

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

7 0
2 years ago
How many moles of carbon are<br> in 7.87x10' carbon atoms?
larisa [96]

Answer:

1.306 moles of C

Explanation:

6 0
1 year ago
Uranium–232 has a half–life of 68.9 years. A sample from 206.7 years ago contains 1.40 g of uranium–232. How much uranium was or
givi [52]

<u>Answer:</u> The initial amount of Uranium-232 present is 11.3 grams.

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

We are given:

t_{1/2}=68.9yrs

Putting values in above equation, we get:

k=\frac{0.693}{68.9}=0.0101yr^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant = 0.0101yr^{-1}

t = time taken for decay process = 206.7 yrs

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process = 1.40 g

Putting values in above equation, we get:

0.0101yr^{-1}=\frac{2.303}{206.7yrs}\log\frac{[A_o]}{1.40}

[A_o]=11.3g

Hence, the initial amount of Uranium-232 present is 11.3 grams.

4 0
1 year ago
Read 2 more answers
The beta oxidation pathway degrades activated fatty acids (acyl-CoA) to acetyl-CoA, which then enters the citric acid cycle. Add
BabaBlast [244]

Answer:

The correct statements are given below

Explanation:

b Enoyl CoA isomerase an enzyme that converts cis double bonds to trans double bonds in fatty acid metabolism,bypasses a step that reduces Q,resulting in the higher ATP yield.

c Even chain fatty acids are oxidized to acetyl CoA in the beta oxidation pathway.

f The final round of beta oxidation foe a 13 carbon saturated fatty acid yields acetyl CoA and propionyl CoA a three carbon fragment.

3 0
1 year ago
Read 2 more answers
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