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3 years ago

Can you live without chemical reactions

Chemistry

2 answers:

Oliga [24]3 years ago

7 1

Answer:

Explanation:

your body wouldnt be able to repair, move, or do anything for that matter if we didnt consistently have chemical reactions in us.

Marat540 [252]3 years ago

8 0

Nothing would change if chemical reactions were a thing. we can’t live without chemical reactions.

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g, Assuming the precipitate is totally insoluble in water, which aqueous ions will be present in the solution (collected in the

Allushta [10]

Answer:

Cl⁻, Na⁺, OH⁻

Explanation:

The titration is:

CuCl₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaCl(aq)

In solution, before the reaction, the ions are Cu²⁺ and Cl⁻. The addition of NaOH (Na⁺ + OH⁻) produce the precipitation of Cu²⁺ forming Cu(OH)₂(s). When you reach the equivalence point, there is no Cu²⁺ because precipitates completely. All OH⁻ ions reacts when are added but when Cu²⁺ is finished, excess OH⁻ ions still in solution helping to detect the equivalence point.

Thus, ions present after the equivalence point are: Cl⁻, Na⁺ (Don't react, spectator ions), and OH⁻.

3 0

3 years ago

1. Cp is smaller than Hx

andreev551 [17]

1 has a higher ionization dismal aoa

8 0

3 years ago

What would happen to the block if it had a density of 0.500 kg/L and was placed in the same 100.0 L tank of water?

raketka [301]

I’m not sure if there was important information in the question before this one, but the answer based on the info I have is B.

The density of water is 1kg/L. Since the density of the block is less, it will float.

6 0

3 years ago

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The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

Answer: The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

Explanation:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

BTW IT IS SCIENCE

Sliva [168]

The answer is A warm air rises cool air sinks

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3 years ago

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