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aliya0001 [1]
2 years ago
11

How many moles of solute are present in 150mL of 0.30 mol/L NaOH solution ourse

Chemistry
1 answer:
Y_Kistochka [10]2 years ago
6 0

Answer:

0.045 moles of NaOH

Explanation:

When this units are together mol/L, we talk about M, molarity.

This kind of concentration shows the relation between moles of solute which are contained in 1L of solution.

In this case, our NaOH solution is 0.30 M

M = mol/L

Then, M . L = mol

We convert volume to L → 150 mL . 1L /1000 mL = 0.150L

0.150 L . 0.30mol/L = 0.045 moles

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Answer:

Explanation:

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7 0
2 years ago
Balance the following chemical reaction:
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Answer:

what the freak is thisnenejehey

7 0
3 years ago
Read 2 more answers
PLS HELP QUICK ALOTTT OF POINTS
timofeeve [1]

Answer:

\boxed {\boxed {\sf 0.80 \ mol\ F}}

Explanation:

We are asked to find how many moles are in 4.8 × 10²³ fluorine atoms. We convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of fluorine.

We will convert using dimensional analysis and set up a ratio using Avogadro's Number.

\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}

We are converting 4.8 × 10²³ fluorine atoms to moles, so we multiply the ratio by this number.

4.8 \times 10^{23} \ atoms \ F *\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}

Flip the ratio so the units of atoms of fluorine cancel each other out.

4.8 \times 10^{23} \ atoms \ F *\frac { 1 \ mol \ F}{6.022 \times 10^{23} \ atoms \ F}

4.8 \times 10^{23}  *\frac { 1 \ mol \ F}{6.022 \times 10^{23} }

Condense into 1 fraction.

\frac { 4.8 \times 10^{23} }{6.022 \times 10^{23} } \ mol \ F

Divide.

0.7970773829 \ mol \ F

The original measurement of atoms has 2 significant figures, so our answer must have the same. For the number we found, that is the hundredths place. The 7 in the thousandths tells us to round the 9 in the hundredths place up to a 0. Then, we also have to round the 7 in the tenths place up to an 8.

0.80 \ mol \ F

4.8 × 10²³ fluorine atoms are equal to <u>0.80 moles of fluorine.</u>

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3 years ago
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<span> ΔT(freezing point)  = (Kf)mi
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3 =3.72m
m = 0.81 mol/kg</span>

4 0
3 years ago
Read 2 more answers
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