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qaws [65]
3 years ago
5

If the density of a 45.0 cm3 block of wood is 0.65 g/ml what is the mass of the wood?

Chemistry
1 answer:
wlad13 [49]3 years ago
5 0

Answer:

\boxed {\tt mass=29.25 \ grams}

Explanation:

The density formula is mass over volume.

d=\frac{m}{v}

Rearrange the formula for the mass, m. Multiply both sides of the formula by v.

d*v=\frac{m}{v}*v

d*v=m

Mass can be found by multiplying the density and volume. The density is 0.65 grams per milliliter and the volume is 45.0 cubic centimeters.

  • A cubic centimeter is equal to a milliliter.
  • Therefore, 45 cubic centimeters also equals 45 milliliters.

d= 0.65 \ g/mL\\v= 45 \ mL

Substitute the values into the formula.

0.65 \ g/mL * 45 \ mL=m

Multiply. Note the milliliters, or mL will cancel out.

0.65 \ g * 45=m

29.25 \ g=m

The mass of the wood is 29.25 grams.

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When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of
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Answer:

Mass of carbon dioxide produced = 52.8 g

Explanation:

Given data:

Mass of carbon react = 14.4 g

Mass of oxygen = 56.5 g

Mass of oxygen left = 18.1 g

Mass of carbon dioxide produced = ?

Solution:

C + O₂     →      CO₂

Number of moles of C:

Number of moles = mass/ molar mass

Number of moles = 14.4 g/ 12 g/mol

Number of moles = 1.2 mol

18.1 g of oxygen left it means carbon is limiting reactant.

Now we will compare the moles of C with CO₂.

                       C             :         CO₂

                        1             :          1

                      1.2           :          1.2

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 1.2  mol × 44 g/mol

Mass = 52.8 g

8 0
3 years ago
An atom has 13 protons, 13 neutrons, and 13 electrons. Another atom has 13 protons, 14 neutrons, and 13 electrons. Analyze and e
olga nikolaevna [1]
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According to Newton's 1st law of motion, what is required to make an object slow down?
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Newton’s first law state that an object in motion stays in motion unless acted upon force so the answer would be RESISTANCE
7 0
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What mass of calcium carbonate is produced when 250 mL of 6.0 M sodium carbonate is added to 750 mL of 1.0 M calcium fluoride
Savatey [412]

<u>Given:</u>

Volume of Na2CO3 = 250 ml = 0.250 L

Molarity of Na2CO3 = 6.0 M

Volume of CaF2 = 750 ml = 0.750 L

Molarity of CaF2 = 1.0 M

<u>To determine:</u>

The mass of CaCO3 produced

<u>Explanation:</u>

Na2CO3 + CaF2 → CaCO3 + 2NaF

Based on the reaction stoichiometry:

1 mole of Na2CO3 reacts with 1 moles of Caf2 to produce 1 mole of caco3

Moles of Na2CO3 present = V * M = 0.250 L * 6.0 moles/L = 1.5 moles

Moles of CaF2 present = V* M = 0.750 * 1 = 0.750 moles

CaF2 is the limiting reagent

Thus, # moles of CaCO3 produced = 0.750 moles

Molar mass of CaCO3 = 100 g/mol

Mass of CaCO3 produced = 0.750 moles * 100 g/mol  = 75 g

Ans: Mass of CaCO3 produced = 75 g

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