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2 years ago

If the mass of a material is 78 grams and the volume of the material is 23 cm3, what would the density of the material be?

Chemistry

2 answers:

Paraphin [41]2 years ago

8 0

Density = mass / volume

Density = (78g) / (23cm^3)

Density = 3.4g/cm^3

Orlov [11]2 years ago

6 0

By definition, density is mass per volume.

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

Plugging in the given values gives:

$d=\frac{78 g}{23 cm^3} =3.3913\frac{g}{cm^3}$

The only operation done in the problem is division, and both numbers have 2 significant figures, so this would be $3.4\frac{g}{cm^3}$ with correct significant figures.

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Valence Bond theory predicts that tin will use what hybrid orbitals in Snf5 -1

LUCKY_DIMON [66]

Answer:

sp3d

Explanation:

The ground state electronic configuration of tin is written as; [Kr] 5s²4d¹⁰5p². Hybridization is a concept used to explain the combination of orbitals of appropriate energy to produce suitable orbitals that could be used for bonding.

In forming the compound Snf5^ -1, we have to hybridize the following orbitals on tin; 5p, 5d and 6s orbitals. This gives us a set of sp3d hybrid hence the answer.

6 0

1 year ago

To test the effectiveness of a gunpowder mixture, 1 gram was exploded under controlled STP conditions, and the reaction chamber

Alekssandra [29.7K]

Answer:

1.3 × 10³ cm³

Explanation:

The gas occupies a volume of V₁ = 310 cm³ under standard temperature and pressure (STP), that is, T₁ = 273.15 K and P₁ = 1.0 atm. In order to find the volume V₂ under different conditions we can use the combined gas law formula.

$\frac{P_{1}.V_{1}}{T_{1}} =\frac{P_{2}.V_{2}}{T_{2}} \\V_{2}=\frac{P_{1}.V_{1}.T_{2}}{T_{1}.P_{2}}=\frac{1.0atm\times 310cm^{3} \times 2473K }{273.15K \times 2.1atm} =1.3 \times 10^{3} cm^{3}$

7 0

2 years ago

The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of

timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = $1.83g/cm^3=1.83g/ml$

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml$

Now we have to calculate the molarity of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L$

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg$

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

7 0

2 years ago

The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g

ra1l [238]

Answer:

1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.

2) When 10.4 grams of carbon monoxide(g) react with excess water(l), 1.04 kJ of energy are absorbed.

Explanation:

1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).

2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ

The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:

$6.97g.\frac{1mol}{22.98g} .\frac{-369kJ}{2mol} =-56.0kJ$

2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).

CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g) ΔH = 2.80 kJ

The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is:

$10.4g.\frac{1mol}{28.01g} .\frac{2.80kJ}{mol} =1.04kJ$

3 0

3 years ago

Which two from this group have an identical number of energy levels?

storchak [24]

Answer: C

Explanation: Took the test

5 0

2 years ago