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mote1985 [20]
3 years ago
13

A sample of an ionic compound that is often used as a dough conditioner is analyzed and found to contain 4.628 g of potassium, 9

.457 g of bromine, and 5.681 g of oxygen.
What is the empirical formula for this compound?

What is it chemical name?
Chemistry
1 answer:
Roman55 [17]3 years ago
5 0

Answer:

The empirical formula of the compound is = KBrO_3

The name of the compound is potassium bromate.

Explanation:

Mass of potassium = 4.628 g

Moles of potassium = \frac{4.628 g}{39 g/mol}=0.1187 mol

Mass of bromine = 9.457 g

Moles of bromine = \frac{9.457 g}{80 g/mol}=0.1182 mol

Mass of oxygen = 5.681 g

Moles of oxygen = \frac{5.681 g}{16 g/mol}0.3551

For empirical formula of the compound, divide the least number of moles from all the moles of elements present in the compound:

Potassium :

\frac{0.1187 mol}{0.1182 mol}=1.0

Bromine;

\frac{0.1182 mol}{0.1182 mol}=1.0

Oxygen ;

\frac{0.3551 mol}{0.1182 mol}=3.0

The empirical formula of the compound is = KBrO_3

The name of the compound is potassium bromate.

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8 0
3 years ago
A blacksmith heated an iron bar to 1445 °C. The blacksmith then tempered the metal by dropping it into 42,800 mL of
Wittaler [7]

Answer:

6626 g

Explanation:

Given that:

Density of water = 1.00 g/ml, volume of water = 42800 ml.

Since density = mass/ volume

mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g

Initial temperature of water = 22°C and final temperature of water = 45°C.

specific heat capacity for water = 4.184 J/g°C

ΔT water = 45 - 22 = 23°C

For iron:

mass = m,  

specific heat capacity for iron  = 0.444 J/g°C

Initial temperature of iron = 1445°C and final temperature of water = 45°C.

ΔT iron = 45 - 1445 = -1400°C

Quantity of heat (Q) to raised the temperature of a body is given as:

Q = mCΔT

The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.

Q water (gain) + Q iron (loss) = 0

Q water = - Q iron

42800 g ×  4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C

m = 4118729.6/621.6

m = 6626 g

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