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Aleks [24]
4 years ago
13

Complete the formula. density = _____ weight/volume weight/length mass/length mass/volume

Chemistry
2 answers:
viva [34]4 years ago
6 0

Answer: Option (d) is the correct answer.

Explanation:

Density is defined as mass per unit volume.

Mathematically,            d = \frac{m}{v}

where                     d = density

                              m = mass

                               v = volume

Density is also known as degree of compactness, that is, more closely the atoms are linked more will be the mass of object or atom. Hence, more will be its density.

Thus, we can conclude that density = \frac{mass}{volume}

faltersainse [42]4 years ago
4 0
Mass/volume cause you divide to get density
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1. A sample of gold (Au) has a mass of 35.12 g.
Alecsey [184]

1a) Answer is: the number of moles of gold (Au) in the sample is 0.178 mol.

m(Au) = 35.12 g; mass of gold.

M(Au) = 196.97 g/mol; molar mass of gold.

n(Au) = m(Au) ÷ M(Au).

n(Au) = 35.12 g ÷ 196.97 g/mol.

n(Au) = 0.178 mol; amount of gold.

1b) Answer is: the number of atoms of gold (Au) is 1.073·10²³.

N(Au) = n(Au) · Na.

N(Au) = 0.178 mol · 6.022·10²³ 1/mol.

N(Au) = 1.073·10²³.

2a) Answer is: 0.0035 moles of sucrose.

m(C₁₂H₂₂O₁₁) = 1.202 g; mass of sucrose.

M(C₁₂H₂₂O₁₁) = 12 · Ar(C) + 22 · Ar(H) + 11 · Ar(O) · g/mol.

M(C₁₂H₂₂O₁₁) = 12 · 12.01 + 22 · 1.01 + 11 · 16 · g/mol.

M(C₁₂H₂₂O₁₁) = 342.3 g/mol; molar mass of sucrose.

n(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).

n(C₁₂H₂₂O₁₁) = 1.202 g ÷ 342.3 g/mol.

n(C₁₂H₂₂O₁₁) = 0.0035 mol; amount of sucrose.

2b) n(C) = 12·n(C₁₂H₂₂O₁₁).

n(C) = 12 · 0.0035 mol.

n(C) = 0.042 mol; amount of carbon in sucrose.

n(H) = 22·n(C₁₂H₂₂O₁₁).

n(H) = 22 · 0.0035 mol.

n(H) = 0.077 mol; amount of hydrogen in sucrose.

n(O) = 11·n(C₁₂H₂₂O₁₁).

n(O) = 11 · 0.0035 mol.

n(O) = 0.0385 mol; amount of oxygen atoms in sucrose.

2c) N(C) = n(C) · Na.

N(C) = 0.042 mol · 6.022·10²³ 1/mol.

N(C) = 2.53·10²²; number of carbon atoms in sucrose.

N(H) = n(H) · Na.

N(H) = 0.077 mol · 6.022·10²³ 1/mol.

N(H) = 4.63·10²²; number of hydrogen atoms in sucrose.

N(O) = n(O) · Na.

N(O) = 0.0385 mol · 6.022·10²³ 1/mol.

N(O) = 2.31·10²²; number of oxygen atoms in sucrose.

Na is Avogadro constant.

4 0
3 years ago
Plz help me I swear to God i will mark you brainiest
Marat540 [252]
I think it’s b hope that helped
3 0
3 years ago
Is a caterpillar a consumer
KonstantinChe [14]
The answers yes because it consumes leaves.
5 0
4 years ago
A recessive trait will never express itself if a dominant trait is present. A. FALSE B. TRUE
Alchen [17]
I think it would be B
hope this works
4 0
4 years ago
Read 2 more answers
A city in Laguna generates 0.96 kg per capita per day of Municipal Solid Waste (MSW). Makati City in Metro Manila generates 1.9
mamaluj [8]

Answer:

i) amount of MSW generated:

Laguna: 19200 Kg/day

Makati: 38000 Kg/day

ii) number of trucks to collect twice weekly:

Laguna: 3 trucks

Makati: 5 trucks

iii) volume of MSW in tons that enter landfill/week:

Laguna: 147.84 ton/week

Makati: 292 ton/week

Explanation:

i) Laguna: 0.96 Kg person / day of MSW * 20000 = 19200 Kg MSW / day

⇒ Laguna: 19200 Kg/day * ( 7day/ week ) = 134400 Kg/week

Makati: 1.9 Kg person / day of MSW * 20000 = 38000 Kg MSW / day

⇒ Makati: 38000 Kg/day * ( 7day/week ) = 266000 Kg/week

ii)  truck capacity = 4.4 ton * ( Kg / 0.0011 ton ) = 4000 Kg

⇒ quote/day = 4000 Kg * 0.75 = 3000 Kg

⇒ loads/day = 2 * 3000 kg = 6000 Kg

⇒ operate/week = 5 * 6000 Kg = 30000 Kg

∴ Laguna:  number of trucks needed/week= 134400 / 30000  = 4.48 ≅ 5 trucks

⇒ number of trucks to collect twice weekly = 5 / 2 = 2.5 ≅ 3 trucks

∴ Makati : number of trucks needed/week = 266000 / 30000 = 8.86 ≅ 9 trucks

⇒ number of trucks to collect twice weekly = 9 / 2 = 4.5 ≅ 5 trucks  

iii) enter landfill/week:

Laguna: 134400Kg MSW/week * ( 0.0011 ton/Kg ) = 147.84 ton/week

Makati: 266000Kg MSW/week * ( 0.0011 ton/Kg ) = 292 ton/week

4 0
3 years ago
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