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3 years ago

How many atoms are in 165 g of calcium?

1 answer:

4 0

Atomic mass Calcium = 40.078 a.m.u

40.078 g ---------------- 6.02x10²³ atoms
165 g -------------------- ??

165 x ( 6.02x10²³) / 40.078 => 2.47x10²⁴ atoms

hope this helps!

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What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago

Given the following reaction: 2D(g) + 3E(g) + F(g) \longrightarrow⟶ 2G(g) + H(g) When the concentration of D is decreasing by 0.

Zepler [3.9K]

Answer:

Rate of reaction = -d[D] / 2dt = -d[E]/ 3dt = -d[F]/dt = d[G]/2dt = d[H]/dt

The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1

E decreseas 3/2 as fast as G increases = 0.30 M/s

Explanation:

Rate of reaction = -d[D] / 2dt = -d[E]/ 3dt = -d[F]/dt = d[G]/2dt = d[H]/dt

When the concentration of D is decreasing by 0.10 M/s, how fast is the concentration of H increasing:

Given data = d[D]/dt = 0.10 M/s

-d[D] / 2dt = d[H]/dt

d[H]/dt = 0.05 M/s

The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1

When the concentration of G is increasing by 0.20 M/s, how fast is the concentration of E decreasing:

d[G] / 2dt = -d[H]/3dt

E decreseas 3/2 as fast as G increases = 0.30 M/s

5 0

3 years ago

Consider the insoluble compound nickel(II) hydroxide , Ni(OH)2 . The nickel ion also forms a complex with cyanide ions . Write a

natali 33 [55]

Answer: Equilibrium constant for this reaction is $2.8 \times 10^{15}$ .

Explanation:

Chemical reaction equation for the formation of nickel cyanide complex is as follows.

$Ni(OH)_{2}(s) + 4CN^{-}(aq) \rightleftharpoons [Ni(CN)_{4}^{2-}](aq) + 2OH^{-}(aq)$

We know that,

K = $K_{f} \times K_{sp}$

We are given that, $K_{f} = 1.0 \times 10^{31}$

and, $K_{sp} = 2.8 \times 10^{-16}$

Hence, we will calculate the value of K as follows.

K = $K_{f} \times K_{sp}$

K = $(1.0 \times 10^{31}) \times (2.8 \times 10^{-16})$

= $2.8 \times 10^{15}$

Thus, we can conclude that equilibrium constant for this reaction is $2.8 \times 10^{15}$ .

4 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).

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7 0

2 years ago

Calcium carbonate (caco3) is an important component of coral reefs. how many moles are in 98.6 g of caco3? type in your answer u

vovangra [49]

0.986 moles of CaO are formed when 98.60 g of $CaCO_3$ decomposes.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × $10^{23}$ of some chemical unit, be it atoms, molecules, ions, or others.

Number of moles of CaCO3 = $\frac{mass}{molar \;mass}$

Number of moles of CaCO3 = $\frac{98.60 g}{100 g/mol}$

= 0.986 moles

Reaction:

$CaCO_3$ → $CaO + CO_2$

Since the reaction is 1:1:1.

So 0.986 moles of CaO are formed.

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7 0

2 years ago