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3 years ago

How many moles of sodium hydroxide are contained in 65.0 mL of a 2.20 M solution?

1 answer:

7 0

2.20 M means there are 2,20 mol of NaOH in 1 000 mL of solution. We can use this proportion as a conversion factor:

$65\ mL\cdot \frac{2.20\ mol}{1\ 000\ mL} = \bf 0.143\ mol$

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how many moles of potassium hydroxide are needed to completely react with 1.73 miles of aluminum sulfate according to the follow

Masteriza [31]

<h3><u>Answer; </u></h3>

=10.38 moles KOH

<h3><u>Explanation</u>;</h3>

The balanced equation.

6KOH + Al2(SO4)3 --> 3K2SO4 + 2Al(OH)3

From the equation;

1 mole of aluminum sulfate requires 6 moles of potassium hydroxide.

Moles of Aluminium sulfate; 1.73 moles

Moles of KOH;

1 mol Al2(SO4)3 : 6 mol KOH = 1.73 mol Al2(SO4)3 : x mol KOH

Thus; x = (6 × 1.73)

<u> =10.38 moles KOH </u>

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3 years ago

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An ionic compound is

Musya8 [376]

Answer:

a compound

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Mass is measured in ____________, and weight is measured in ______________. A) kilograms, grams B) grams, kilograms C) newtons,

frez [133]

I'm pretty sure the answer is letter D.

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Which would affect global transportation if depleted?

Elis [28]

Awnser: C. petrolium

See, petroleum (i.e. oil) is transformed into gasoline.

"Sometimes, petroleum and crude oil are used to mean the same thing, but petroleum itself is a broad range of petroleum products including crude oil itself. We use the term 'petroleum products after crude oil is refined in a factory." - <span>www.eschooltoday.com/energy/non-renewable-energy/what-is-petroleum.html</span>

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3 years ago

How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci

bezimeni [28]

Answer:

Approximately $0.08\; \rm mol$ , assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure: $25\; \rm ^{\circ}C$ and $P = 101.325 \; \rm kPa$ .

The question states that the volume of this gas is $V = 2\; \rm dm^{3}$ .

Convert the unit of all three measures to standard units:

$\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}$ .

$\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}$ .

$\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}$ .

Look up the ideal gas constant in the corresponding units: $R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}$ .

Let $n$ denote the number of moles of this gas in that $V = 2\; \rm dm^{3}$ . By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

$P \cdot V = n \cdot R \cdot T$ .

Rearrange this equation and solve for $n$ :

$\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}$ .

In other words, there is approximately $2\; \rm mol$ of this gas in that $V = 2\; \rm dm^{3}$ .

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3 years ago