Theory that the lithosphere is divided into tectonic plates that slowly move on

Caclulate the pH for a solution with an [H+] = 9.75 x10-4

vovangra [49]

Answer:

pH=3.01

Explanation:

pH=-log[H+]

=-log(9.74x10e-4)

=3.01

3 0

3 years ago

Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of chromium(III) bromide and pota

expeople1 [14]

Explanation:

An ionic equation will be the one in which all the participating species will be present as ions.

The given reaction will be as follows.

$CrBr_{3} + K_{2}CO_{3} \rightarrow Cr_{2}(CO_{3})_{3}(s) + KBr$

Balancing this equation by multiplying $CrBr_{3}$ by 2 and $K_{2}CO_{3}$ by 3 on reactant side. Whereas multiply KBr by 6 on product side.

$2CrBr_{3} + 3K_{2}CO_{3} \rightarrow Cr_{2}(CO_{3})_{3}(s) + 6KBr$

Hence, the net ionic equation will be as follows.

$2Cr^{3+}(aq) + 3CO^{2-}_{3}(aq) + 6K^{+} + 6Br^{-} \rightarrow Cr_{2}(CO_{3})_{3}(s) + 6K^{+} + 6Br^{-}$

As both $K^{+}$ and $Br^{-}$ are spectator ions. Hence, the net ionic equation will be as follows.

$2Cr^{3+}(aq) + 3CO^{2-}_{3}(aq) \rightarrow Cr_{2}(CO_{3})_{3}(s)$

4 0

2 years ago

Is salad biotic or abiotic

Elenna [48]

Salad is biotic, it is made from plants and plants are alive.
biotic means alive
abiotic means not alive

5 0

3 years ago

For the reaction of sodium bromide with chlorine gas

lina2011 [118]

The reaction of sodium bromide with chlorine gas is Cl₂(aq) + 2Na + 2Br? 2Na + 2Cl⁻ + Br²(aq).

<h3>What is sodium bromide?</h3>

Sodium bromide is an inorganic compound, white, crystalline with high melting point.

The reaction between halogens is redox reaction

Oxidation – 2Br⁻ ? Br₂ + 2e⁻ loss of electron.

Reduction – Cl₂ + 2e⁻ ? 2Cl⁻ gains of electron.

Thus, the correct option is Cl₂(aq) + 2Na + 2Br? 2Na + 2Cl⁻ + Br²(aq).

Learn more about sodium bromide

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7 0

1 year ago

At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction:

Butoxors [25]

<u>Answer:</u> The equilibrium concentration of ICl is 0.27 M

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.45 moles

Initial moles of chlorine gas = 0.45 moles

Volume of the flask = 2.0 L

The molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}$

Initial concentration of iodine gas = $\frac{0.45}{2}=0.225M$

Initial concentration of chlorine gas = $\frac{0.45}{2}=0.225M$

For the given chemical equation:

$2ICl(g)\rightarrow I_2(g)+Cl_2(g);K_c=0.11$

As, the initial moles of iodine and chlorine are given. So, the reaction will proceed backwards.

The chemical equation becomes:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g);K_c=\frac{1}{0.11}=9.091$

<u>Initial:</u> 0.225 0.225

<u>At eqllm:</u> 0.225-x 0.225-x 2x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[ICl]^2}{[Cl_2][I_2]}$

Putting values in above equation, we get:

$9.091=\frac{(2x)^2}{(0.225-x)\times (0.225-x)}\\\\x=0.135,0.668$

Neglecting the value of x = 0.668 because equilibrium concentration cannot be greater than the initial concentration

So, equilibrium concentration of ICl = 2x = (2 × 0.135) = 0.27 M

Hence, the equilibrium concentration of ICl is 0.27 M

6 0

2 years ago