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spin [16.1K]
3 years ago
11

What is the effect of temperature and density on the movement of water?

Chemistry
1 answer:
Rus_ich [418]3 years ago
8 0
<span>temperature affects the movement of water because if the liquid is cold the molecules come together. Therefore, the water moves slower, and if the liquid is hot the molecules spread apart and move faster.<span>
</span></span>
You might be interested in
Zinc sulfate is a 2-ion electrolyte,
Gekata [30.6K]

<u>Answer:</u> The value of <em>i</em> is 1.4 and 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.

<u>Explanation:</u>

The equation used to calculate the Vant' Hoff factor in dissociation follows:

\alpha =\frac{i-1}{n-1}

where,

\alpha = degree of dissociation = 40% = 0.40

i = Vant' Hoff factor

n = number of ions dissociated = 2

Putting values in above equation, we get:

0.40=\frac{i-1}{2-1}\\\\0.40=i-1\\\\i=1.4

The equation used to calculate the degee of dissociation follows:

\alpha =\frac{\text{Number of particles dissociated}}{\text{Total number of particles taken}}

Total number of particles taken = 100

Degree of dissociation = 40% = 0.40

Putting values in above equation, we get:

0.40=\frac{\text{Number of particles dissociated}}{100}\\\\\text{Number of particles dissociated}=(0.40\times 100)=40

This means that 40 particles are dissociated and 60 particles remain undissociated in the solution.

Hence, 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.

7 0
2 years ago
Consider the reaction
SOVA2 [1]

Answer :

(a) The average rate will be:

\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s

(b) The average rate will be:

\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)

The expression for rate of reaction :

\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}

\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}

\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}

\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}

\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}

Thus, the rate of reaction will be:

\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}

<u>Part (a) :</u>

<u>Given:</u>

\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s

As,  

-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}

and,

\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}

\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s

\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s

<u>Part (b) :</u>

<u>Given:</u>

\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s

As,  

-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}

and,

-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}

\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s

\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s

5 0
3 years ago
In your own words, please describe the difference between a complete ionic equation and a net ionic equation.
beks73 [17]

Explanation:

In the molecular equation for a reaction, all of the reactants and products are represented as neutral molecules (even soluble ionic compounds and strong acids). In the complete ionic equation, soluble ionic compounds and strong acids are rewritten as dissociated ions.

The net ionic equation is a chemical equation for a reaction that lists only those species participating in the reaction. The net ionic equation is commonly used in acid-base neutralization reactions, double displacement reactions, and redox reactions.

6 0
3 years ago
g A radioactive isotope of mercury, 197Hg, decays to gold, 197Au, with a disintegration constant of 0.0108hrs.-1. What % of the
weqwewe [10]

Answer:

7.49% of Mercury

Explanation:

Let N₀ represent the original amount.

Let N represent the amount after 10 days.

From the question given above, the following data were obtained:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Next, we shall convert 10 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

10 days = 10 day × 24 h / 1 day

10 days = 240 h

Thus, 10 days is equivalent to 240 h.

Finally, we shall determine the percentage of Mercury remaining as follow:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Log (N₀/N) = kt /2.303

Log (N₀/N) = 0.0108 × 240 /2.303

Log (N₀/N) = 2.592 / 2.303

Log (N₀/N) = 1.1255

Take the anti log of 1.1255

N₀/N = anti log 1.1255

N₀/N = 13.3506

Invert the above expression

N/N₀ = 1/13.3506

N/N₀ = 0.0749

Multiply by 100 to express in percent.

N/N₀ = 0.0749 × 100

N/N₀ = 7.49%

Thus, 7.49% of Mercury will be remaining after 10 days

5 0
2 years ago
A 500 gram piece of metal had a volume of 2.75cm3. What is it’s density
beks73 [17]

Answer:

181.82 g/cm3

Explanation:

density is mass / volume so it is 500 / 2.75=181.82 g/cm3

8 0
3 years ago
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