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Anton [14]
3 years ago
9

Calculate the number of moles of Californium represented by 5.92 x 1024 atoms of Californium. Enter your answer in the provided

box. mol Cf
Chemistry
1 answer:
Fed [463]3 years ago
8 0

Answer:

9.834 moles Cf.

Explanation:

The number of moles of a substance is an easy way to represents its amount. Avogadro has determined that the total amount in 1 mol is equal to 6.02x10²³(Avgadros' number), so 1 mol has 6.02x10²³ atoms, molecules, ions, or what we are measuring. So:

1 mol of Cf -------------------- 6.02x10²³ atoms

x -------------------- 5.92x10²⁴

By a simple direct three rule:

6.02x10²³x = 5.92x10²⁴

x = 5.92x10²⁴/6.02x10²³

x = 9.834 moles Cf

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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
When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2
Arlecino [84]

Answer:

1.62

Explanation:

From the given information:

number of moles of benzamide  =\dfrac{70.4 \ g}{121.14 \ g/mol}

= 0.58 mole

The molality = \dfrac{mass \ of \ solute (i.e. \ benzamide )}{mass \ of \ solvent  }

= \dfrac{0.58 }{0.85 }

= 0.6837

Using the formula:

\mathbf {dT  = l   \times  k_f  \times m}

where;

dT = freezing point = 27

l = Van't Hoff factor = 1

kf = freezing constant of the solvent

∴

2.7 °C = 1 × kf ×  0.6837 m

kf = 2.7 °C/ 0.6837m

kf = 3.949 °C/m

number of moles of NH4Cl = \dfrac{70.4 \ g}{53.491 \  g /mol}

= 1.316 mol

The molality = \dfrac{1.316 \ mol}{0.85 \ kg}

= 1.5484

Thus;

the above kf value is used in determining the  Van't Hoff factor for  NH4Cl

i.e.

9.9 = l × 3.949 × 1.5484 m

l = \dfrac{9.9}{3.949 \times 1.5484 \ m}

l = 1.62

5 0
2 years ago
A sample of magnesium oxide has a mass of 94.4 g. How many molecules
ella [17]

Answer: 1.414x10^24 molecules in 94.4g MgO

Explanation: molar mass MgO 40.204

molecules in 40.204 g MgO = avogadro number

molecules in 94.4 g MgO = (94.4/40.204)*avogadro number

(94.4/40.204)*6.02214076*10^23 = 14.14x10^23

7 0
3 years ago
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