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2 years ago

5) What is the volume of 6.70 x 10^15 molecules of propane, C3 H8 at STP?

Chemistry

1 answer:

lisabon 2012 [21]2 years ago

4 0

Divide the number by 6.02 x 10^23 so the measurement is in moles and multiply by 22.4 to convert it into liters

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Help plzzzzz ASAP!!!!!!!!

SIZIF [17.4K]

1. NaF, Na₂S, Na₃P, Na₂O

2. MgF₂, MgS, Mg₃P₂, MgO

3. AlF₃, Al₂S₃, AlP, Al₂O₃

<h3>Further explanation</h3>

Given

Ionic charge

Required

The formula of binary ionic compounds

Solution

Ionic compounds consisting of cations (ions +) and anions (ions -)

Ionic compounds usually consist of metal cations and non-metal anions

Metal: cation, positively charged.

Nonmetal: negatively charged

The anion cation's charge is crossed

The ionic compounds :

1. NaF, Na₂S, Na₃P, Na₂O

2. MgF₂, MgS, Mg₃P₂, MgO

3. AlF₃, Al₂S₃, AlP, Al₂O₃

4 0

2 years ago

A bike lost 20% of its value after the first year, its value after the first year is $1039. Calculate the original price of the

musickatia [10]

Answer:

$1246.90

Explanation:

Since the bike lost $20%$ % of it's value and it now currently at $ $$1039$ , we have to do 20% * $1039 to find the amount of money lost. 20%*1039=207.8. We have to add it up to find the original value so 1039+207.8=$1246.8

6 0

2 years ago

Acetic acid only partially ionizes in water. A solution contains a large quantity of acetic acid dissolved in water. How can thi

dusya [7]

B because acetic acid is a weak acid and large quantity means you make it become concentrated

4 0

3 years ago

Read 2 more answers

A compound with the empirical formula CH 2 O has a formula mass of 180 g/mol. What is its molecular formula

olga nikolaevna [1]

Empirical formula mass

12+2(1)+16
28+2
30g/mol

Molecular fornula mass:-180g/mol

n=Molecular formula mass/Empirical formula mass
m=180/30
n=6

Molecular formula:-

n×Empirical formula
6(CH_2O
C_6H_12 O_6

5 0

2 years ago

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