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Amanda [17]
3 years ago
13

All living organisms are made out of cells. Select three cell parts that all cells contain.

Chemistry
1 answer:
algol [13]3 years ago
5 0

Answer:

cell membrane, ribosomes and dna..?

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If gas particles start colliding with the walls of their metallic container with increased force, what is their direct effect? l
stepladder [879]
The answer is B) Higher gas pressure.
8 0
3 years ago
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Consider the following reaction where K. = 154 at 298 K: 2NO(g) + Brz(9) 2NOBr(g) A reaction mixture was found to contain 2.69x1
bekas [8.4K]

Explanation:

2NO(g) + Br_2(g)\rightleftharpoons 2NOBr(g)

Equilibrium constant of reaction = K=154

Concentration of NO = [NO]=\frac{2.69\times 10^{-2} mol}{1 L}=2.69\times 10^{-2} M

Concentration of bromine gas = [Br_2]=\frac{3.85\times 10^{-2} mol}{1 L}=3.85\times 10^{-2} M

Concentration of NOBr gas = [Br_2]=\frac{9.56\times 10^{-2} mol}{1 L}=9.56\times 10^{-2} M

The reaction quotient is given as:

Q=\frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{(9.56\times 10^{-2} M)^2}{(2.69\times 10^{-2} M)^2\times 3.85\times 10^{-2} M}

Q=328.06

Q>K

The reaction will go in backward direction in order to achieve an equilibrium state.

1. In order to reach equilibrium NOBr (g) must be produced.  False

2. In order to reach equilibrium K must decrease. False

3. In order to reach equilibrium NO must be produced. True

4. Q. is less than K . False

5. The reaction is at equilibrium. No further reaction will occur. False

8 0
3 years ago
The ________ orbital is degenerate with 5py in a many-electron atom.
arsen [322]

The ________ orbital is degenerate with 5py in a many-electron atom.


<h2>5px is the correct answer</h2>
5 0
3 years ago
The value of the solubility product constant for Ag2CO3 is 8.5 × 10‒12 and that of Ag2CrO4 is 1.1 × 10‒12. From this data, what
Lena [83]

Answer:

B) 7.7

Explanation:

For the reaction    Ag2CO3(s) + CrO42‒(aq) → Ag2CrO4(s) + CO32‒(aq)

Kc = (CO₃²⁻) / (CrO₄²⁻)

and the Ksp given are

Ag₂CO₃    ⇒  2 Ag⁺(aq) + CO₃²⁻(aq)    Ksp₁ = (Ag⁺)²(CO₃²⁻)  

Ag₂CrO₄   ⇒  2 Ag⁺(aq)+ CrO₄²⁻(aq)   Ksp₂ = (Ag⁺)²(CrO₄²⁻)

Where (...) indicate concentrations M

Notice if we divide the expressions for Ksp we get:

Ksp₁/Ksp₂ = (CO₃²⁻)  / (CrO₄²⁻) = 8.5 x 10⁻¹² / 1.1 x 10⁻¹² = 7.7

which is the desired answer.

7 0
3 years ago
G. Amount of charge required to reduce
Nady [450]

Answer:

\boxed{\text{c) 4 F}}

Explanation:

1. Write the skeleton equation for the half-reaction

NO₃⁻ ⟶ N₂O

2. Balance all atoms other than H and O

2NO₃⁻ ⟶ N₂O

3. Balance O by adding H₂O molecules to the deficient side.

2NO₃⁻ ⟶ N₂O + 5H₂O

4. Balance H by adding H⁺ ions to the deficient side.

2NO₃⁻ + 10H⁺ ⟶ N₂O + 5H₂O

5. Balance charge by adding electrons to the deficient side.

2NO₃⁻ + 10H⁺ + 8e⁻ ⟶ N₂O + 5H₂O

The amount of charge required to reduce 2 mol of NO₃⁻ is 8 F

\text{The amount of charge required to reduce 1 mol of NO$_{3}^{-}$ is \boxed{\textbf{4 F}}}

4 0
3 years ago
Read 2 more answers
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