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3 years ago

7

Which factors would decrease the rate of a reaction? I. Lowering the temperature II. Increasing the concentration of reactants I

II. Adding a catalyst

1 answer:

Basile [38]3 years ago

6 0

I. Lowering the temperature would decrease the rate of the reaction.

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In pure water at 25 °C, the concentration of a saturated solution of CuF2 is 7.4 × 10−3 M. If measured at the same temperature,

Romashka-Z-Leto [24]

Answer:

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.

Explanation:

Consider the ICE take for the solubility of the solid, CuF₂ as:

CuF₂ ⇄ Cu²⁺ + 2F⁻

At t=0 x - -

At t =equilibrium (x-s) s 2s

The expression for Solubility product for CuF₂ is:

$K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2$

$K_{sp}=s\times {2s}^2$

$K_{sp}=4s^3$

Given s = 7.4×10⁻³ M

So, Ksp is:

$K_{sp}=4\times (7.4\times 10^{-3})^3$

$K_{sp}=4\times (7.4\times 10^{-3})^3$

Ksp = 1.6209×10⁻⁶

Now, we have to calculate the solubility of CuF₂ in NaF.

Thus, NaF already contain 0.20 M F⁻ ions

Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

CuF₂ ⇄ Cu²⁺ + 2F⁻

At t=0 x - 0.20

At t =equilibrium (x-s') s' 0.20+2s'

The expression for Solubility product for CuF₂ is:

$K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2$

$1.6209\times 10^{-6}={s}'\times ({0.20+2{s}'})^2$

Solving for s', we get

<u>s' = 4.0×10⁻⁵ M</u>

<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.</u>

3 0

4 years ago

Chlorine consists of two isotopes with masses of 35 and 37. If the average atomic mass of a sample of chlorine atoms is 35.5, wh

Elza [17]

Answer:

The ratio 35Cl/37Cl = 3/1

Explanation:

<u>Step 1:</u> Data given

Chlorine has 2 isotopes:

- mass = 35 g/mol

- mass = 37 g/mol

Average molar mass of chlorine = 35.5 grams

<u>Step 2: </u>Calculate the % of isotopes

35x + 37y = 35.5

x+y = 1 or x = 1-y

35(1-y) + 37y = 35.5

35-35y +37y = 35.5

0.5 = 2y

y = 0.25 = 37Cl

x = 1 - 0.25 = 0.75 = 35Cl

<u>Step 3: </u>

The ratio 35Cl/37Cl = 0.75/0.25 = 3/1

7 0

4 years ago

H2SO4(aq)+NaOH(s)> (neutralization of both acidic protons)

RUDIKE [14]

H2SO4(aq)+NaOH(s)----->NA2SO4 + H20

3 0

3 years ago

On the basis of le chatelier's principle explain why ag2co3 dissolves when hno3 is added

Bogdan [553]

The reaction is an equilibrium represented by the equation

<span>Ag2CO3(s) + 2 HNO3(aq) <----> 2 AgNO3(aq) + H2O(l) + CO2(g) </span>

From the <span>Le Chatelier's Principle which </span>states that changing a factor such as concentration, temperature, or pressure of a reaction at equilibrium will cause the reaction to shift in the direction that counteracts the effect of that change.
Therefore; the CO2 produced starts escaping and the concentration and pressure of CO2 drops. The system responds by trying to increase the concentration and pressure of CO2 by producing more. This means more and more Ag2CO3 will dissolve due to reaction with the acid, HNO3.
This continues until one of the reactants is exhausted.

7 0

3 years ago

Read 2 more answers

A rock is found to contain 2.2 grams of uranium-235, a radioactive isotope. If the rock has undergone 2.75 half-lives, what was

Verizon [17]

Answer:

=14.8 grams

Explanation:

The remaining amount is normally calculated using the formula:

Remaining mass= 1/2ⁿ × Original mass where n is the number of half-lives.

Therefore, original mass= Remaining mass × 2ⁿ

Remaining mass= 2.2 grams

Number of half lives= 2.75 half lives

Original mass= 2.2g × 2²·⁷⁵

=14.8 grams

3 0

3 years ago