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3 years ago

A sample of iron has a mass of 46.8 grams and a volume of 6 cm3. what is the density of iron?

Chemistry

2 answers:

Dovator [93]3 years ago

6 0

7.8 g/cm3 because Mass/Volume=Density

eduard3 years ago

5 0

Answer:

$7.8 g/cm^{3}$

Explanation:

Density, denoted by the greek symbol rho (ρ), is the mass per unit of volume and it is also called specific mass. For a pure substance, density has the same value as the mass concentration.

The equation to calculate the density is as follows:

ρ= $\frac{m}{v} =\frac{mass}{volume}$

Replacing the mass and volume given in the statement, we optain:

ρ= $\frac{46.8 g}{6cm^{3} } =7.8\frac{g}{cm^{3} }$

Obtaining a similar value to that of the theoretical density of iron, wich is $7.874 g/cm^{3}$

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Nadusha1986 [10]

One of the basic building blocks of matter, which cannot be broken down by chemical means is an element.

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Other example, krypton is a chemical element with symbol Kr and atomic number 36.

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2 years ago

determine the freezing point depression of a solution that contains 30.7 g glycerin (c3h8o3, molar mass

Blababa [14]

The freezing point depression of a solution containing 30.7 g of glycerin is calculated as -1.65°C

Equating :

It is given that,

Given mass of glycerin is = 30.7 grams (Solute)

Volume of water = 376 mL

$K_{f}$ or molar -freezing-depression point is = 1.86°C/m

Molar mass of glycerin = 92.09 g/mole

Now, to work out the value, the mass of water should be known. Thus, to calculate, the formula used will be:

Mass = Density X Volume

Mass = 1.0 g/mL X 376 mL

Mass = 376 g or 0.376 Kg

Using the formula of melting point depression, the equation becomes:

Δ $T_{f}$ = i × $K_{f}$ ×m

T⁰- $T_{s}$ = $i *K_{f} *\frac{mass of glycerin}{molar mass of glycerin * mass of water in kg}$

in which,

Δ $T_{f}$ = change in freezing point

Δ $T_{s}$ = freezing point of solution that has to be find

ΔT° = freezing point of water ()

Since, glycerin is a non-electrolyte, the Van't Hoff factor will be 1.

Substituting the values in the above equation:

0⁰C₋T $_{s}$ = 1 ×1.86°C/m × $\frac{30.7}{92.09g/mol * 0.376kg}$

$T_{s}$ = -1.65°C

Thus, the freezing point depression of a solution is -1.65°C

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Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all less than that of the pure solvent and is directly proportional to the molality of the solute

Is melting point elevation or depression?

Boiling point elevation is that the raising of a solvent's boiling point due to the addition of a solute. Similarly, melting point depression is the lowering of a solvent's freezing point due to the addition of a solute. In fact, because the boiling point of a solvent increases, its melting point decreases

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1 year ago

PLEASE ANSWER BOTH QUESTIONS I WILL MARK YOU BRAINLIEST IF IT IS CORRECT!!

seropon [69]

Answer:

1st Question: A

2nd Question: B

Explanation:

The 1st answer would be A because if a sample is at absolute zero then the sample is at its lowest temperature none of the molecules would be able to move, this is because lower temperature= lower kinetic energy.

The 2nd answer would be B because if a sample has more temperature it speeds up it has more temperature and more kinetic energy, meaning it would move faster because there is more temperature.

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3 years ago

Please help me im stuck on this!!

lisov135 [29]

The answer is decomposition

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3 years ago

Read 2 more answers

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kakasveta [241]

Answer: CaSO3

Explanation:Please see attachment for explanation