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ArbitrLikvidat [17]
3 years ago
12

Five mineral samples of equal mass of calcite, caco3 (mm 100.085) , had a total mass of 10.9 ± 0.1 g. what is the average mass o

f calcium in each sample? (assume that the relative uncertainties in atomic mass are small compared the uncertainty of the total mass.)
Chemistry
1 answer:
g100num [7]3 years ago
6 0

The molecular mass of calcite is 100.085 (given).

Atomic mass of calcium = 40.078 amu

Percent amount of calcium in calcite = \frac{40.078}{100.085}\times 100 = 40.044%

The total mass of five mineral samples = 10.9\pm 0.1 g  (given)

So, the mass of one sample = \frac{10.9}{5} = 2.18 g

and \frac{0.1}{5} = 0.02 g

The average mass of calcium in sample = \frac{2.18 g\times 40.044}{100} = 0.873 g

and \frac{0.02 g\times 40.044}{100} = 0.008 g

Hence, the average mass of calcium in each sample is 0.873 g\pm 0.008 g.

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If you push on a wall with 200n with what forth with the wall push back
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Why does a can of diet cola float in water but a can of regular cola does not
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7 0
3 years ago
A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and place
kakasveta [241]

Answer:

Explanation:

Volume of silver cube = 2.42³ = 14.17 cm³

mass of silver cube = volume x density

= 14.17 x 10.49 = 148.64 gm

Volume of gold cube = 2.75³ = 20.8  cm³

mass of gold cube =  20.8 x 19.3 = 401.44 gm

specific heat of silver and gold are .24 and .129 J /g°C

mass of 112 mL water = 112 g

Heat absorbed = heat lost = mass x specific heat x temperature fall or rise

Heat lost by metals

= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )

= (35.67 + 51.78 ) x ( 85.4 - T )

87.45 x ( 85.4 - T )

= 7468.23 - 87.45 T

Heat gained by water

= 112 x 1 x ( T - 20.5 )

= 112 T - 2296

Heat lost = heat gained

7468.23 - 87.45 T = 112 T - 2296

199.45 T = 9764.23

T = 48.95° C

7 0
3 years ago
Given that the molar mass of NaCl is 58.44 g/mol, what is the molarity of a solution that contains 87.75 g of NaCl in 500. mL of
Stells [14]

<u>Given information:</u>

Mass of NaCl (m) = 87.75 g

Volume of solution (V) = 500 ml = 0.5 L

Molar mass of NaCl (M) = 58.44 g/mol

<u>To determine:</u>

The molarity of NaCl solution

<u>Explanation:</u>

Molarity is defined as the number of moles of solute(n) dissolved per liter of solution (V)

i.e. M = moles of solute/liters of solution = n/V

Moles of solute (n) = mass of solute (m)/molar mass (M)

moles of NaCl = 87.75 g/58.55 g.mol-1 = 1.499 moles

Therefore,

Molarity of NaCl = 1.499 moles/0.5 L = 2.998 moles/lit ≅ 3 M

<u>Ans: (D)</u>

4 0
3 years ago
Read 2 more answers
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