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Naddik [55]
3 years ago
12

Explain how you would convert the moles of a compound to the made (grams) of the same compound​

Chemistry
1 answer:
Oduvanchick [21]3 years ago
3 0

Answer:

see notes below

Explanation:

The mole is the mass of substance containing 1 Avogadro's Number of particles. That is, 1 mole substance = 1 formula weight. For elements, 1 mole weight is equal to the atomic weight expressed as grams. For molecules, 1 mole weight is equal to the molecular weight expressed as grams.

1 mole = 1 formula weight

<u>Moles to Grams and Grams to Moles</u>

Grams => Moles

Given grams, moles = mass given / formula weight

*Ask the question => How many formula weights are there in the given mass? => Results is always moles.

Moles => Grams

Given moles,  grams = moles given X formula weight

*Summary

Grams to Moles => divide by formula weight

Moles to Grams => multiply by formula weight

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Phantasy [73]
D
ClO-(aq) + H2O(l) + 2e- Cl-(aq) + 2OH-(aq) apex
6 0
3 years ago
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A liquid occupies 321ml and has a mass of 376g. what is the density of the liquid?​
oee [108]

Answer:

The density of the liquid is 1.153 grams/ml.

Explanation:

You have to use the fomula of density

density = mass/volume ⇒ density = 376 grams/321 ml ⇒

density = 1.153 grams/ml.

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2 years ago
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8 0
3 years ago
Calculate the molarity of the two solutions.
daser333 [38]

1. 0.33 M

2. 0.278 M

<h3>Further explanation</h3>

Molarity is a way to express the concentration of the solution

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

1. 0.350 mol of NaOH in 1.05 L of solution.

n=0.35

V=1.05 L

Molarity :

\tt M=\dfrac{0.35}{1.05}=0.33

2. 14.3 g of NaCl in 879 mL of solution.

mol NaCl(MW=58.5 g/mol) :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{14.3~g}{58.5~g/mol}=0.244

Molarity :

\tt M=\dfrac{0.244}{0.879~L}\\\\M=0.278

4 0
2 years ago
If a pharmacist adds 10 ml of purified water to 30 ml of a solution having a specific gravity of 1.30, calculate the specific gr
xxMikexx [17]

The specific gravity of a sample is the ratio of the density of the sample with respect to one standard sample. The standard sample used in specific gravity calculation is water whose density is 1 g/mL. The solution having specific gravity 1.30 is the density of the sample that is 1.30 g/mL. Thus the weight of the 30 mL sample is (30×1.30) = 39 g.

Now the mass of the 10 mL of water is 10 g as density of water is 10 g/mL. Thus after addition the total mass of the solution is (39 + 10) = 49g and the volume is (30 + 10) = 40 mL. Thus the density of the mixture will be \frac{49}{40}=1.225 g/mL. Thus the specific gravity of the mixed sample will be 1.225 g/mL.

6 0
3 years ago
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