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3 years ago

If the atomic weight of nitrogen is 14.01, what is the mass of the nitrogen atoms in one mole of cadmium nitrate, Cd(NOÀ)

1 answer:

8 0

The answer is 28.02.

The chemical formula for cadmium nitrate is Cd(NO₃)₂.

There are in total 2 nitrogen atoms in Cd(NO₃)₂. If <span>the atomic weight of nitrogen is 14.01</span>, the mass <span>of two nitrogen atoms in one mole of cadmium nitrate is 28.02:
2 </span>· 14.01 = 28.02

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H20+ SO3 ? Which compound is likely produced by the chemical reaction?

shusha [124]

Answer:

h2so4

Explanation:

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3 years ago

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C.Convert the following using the molar mass of each:

Alik [6]

Answer:

1. 1.145 mole

2. 19.368g

3. 66.67 moles

4. 0.0056g

Explanation:

The following calculations gives the answers to the various questions asked above:

1. Molar mass of NaCl = 23 + 35.5 = 58.5g/mol

Mass of NaCl from the question = 67g

Number of mole NaCl =.?

Number of mole = Mass /Molar Mass

Number of mole NaCl = 67/58.5

Number of mole of NaCl = 1.145 mol

2. Number of mole of H2O = 1.076 mole

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O =?

Mass = number of mole x molar Mass

Mass of H2O = 1.076 x 18

Mass of H2O = 19.368g

3. Molar mass of C6H12O6 = (12x6) + (12x1) + (16x6) = 72 + 12 + 96 = 180g/mol

Mass of C6H12O6 = 1.2x104g

Mole of C6H12O6 =.?

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4. Molar mass of Br2 = 2 x 80 = 160g/mol

Mole of Br2 = 3.5x10^-5 moles

Mass of Br2 =?

Mass = number of mole x molar Mass

Mass of Br2 = 3.5x10^-5 x 160

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3 years ago

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in a typical person, the level of glucose is about 85 mg/ 100 mL of blood. if an average body contains about 11 pt of blood, how

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Let's note that 1 pint = 473.1765 mL, so 11 pints should be 5204.9415 mL.

We make a proportion out of the word problem

(85 mg glucose/ 100 mL) times (1 g/ 1000 mg) = 4.4242 grams of glucose

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3 years ago

Estimate the number of moles of water in all the Earth's oceans. Assume water covers 75% of the Earth to an average depth of 3 k

o-na [289]

Answer:

there are approximately n ≈ 10²² moles

Explanation:

Since the radius of the earth is approximately R=6378 km= 6.378*10⁶ m , then the surface S of the earth would be

S= 4*π*R²

since the water covers 75% of the Earth's surface , the surface covered by water Sw is

Sw=0.75*S

the volume for a surface Sw and a depth D= 3 km = 3000 m ( approximating the volume through a rectangular shape) is

V=Sw*D

the mass of water under a volume V , assuming a density ρ= 1000 kg/m³ is

m=ρ*V

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n = m/M

then

n = m/M = ρ*V/M = ρ*Sw*D/M = 0.75*ρ*S*D/M = 3/4*ρ*4*π*R² *D/M = 3*π*ρ*R² *D/M

n=3*π*ρ*R² *D/M

replacing values

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n ≈ 10²² moles

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Answer:

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