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4 years ago

Whuch expression is equal to the number of grams in 2.43 kg

Chemistry

1 answer:

schepotkina [342]4 years ago

7 0

2.43*1000.

There are 1000 grams in a kilogram

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A certain gas is present in a 13.0 l cylinder at 4.0 atm pressure. if the pressure is increased to 8.0 atm , the volume of the g

ella [17]

Boyle's Law is k = PV so Initial k = 13.0 L x 4.0 atm = 52 L atm Final kf = 6.5 L x 8 atm = 52 L atm The gas obeys Boyle's Law The answer with two significant figures separated by a comma is k = 52, kf = 52.

5 0

3 years ago

Read 2 more answers

How many grams are in 2.45E24 formula units of Fp3BZ2? The molar mass of Fp3Bz2. is 97.05 g/mol.

dlinn [17]

Answer:

394.99g

Explanation:

The number of moles of a substance can be calculated by dividing the number of atoms of such substance by Avagadro's number (6.02 × 10^23)

n = nA ÷ 6.02 × 10^23

The number of atoms of Fp3BZ2 in this question is 2.45E24 formula units i.e. 2.45 × 10^24

n = 2.45 × 10^24 ÷ 6.02 × 10^23

n = 2.45/6.02 × 10^(24-23)

n = 0.407 × 10¹

n = 4.07moles

Using mole = mass/molar mass

Where; molar mass of Fp3Bz2. is 97.05

g/mol.

mass = molar mass × mole

mass = 97.05 × 4.07

mass = 394.99g

4 0

3 years ago

Which is a factor in determining the average atomic mass of an element?

kvasek [131]

The factor in determining the average atomic mass of an element is:
B or 2 relative abundance of each isotope because the by looking at how many protons , electrons and neutrons the most isotope is of the element has relative abundance.

7 0

3 years ago

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What type of energy results from the

jeka94

Answer:

C fghhtrsfeagutyrwraqedf

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3 years ago

Nitrogen dioxide is used industrially to produce nitric acid, but it contributes to acid rain and photochemical smog. What volum

zhuklara [117]

Answer:

The volume of nitrogen oxide formed is 35.6L

Explanation:

The reaction of nitric acid with copper is:

Cu(s) + 4HNO₃ → Cu(NO₃)₂ + 2NO₂(g) + 2H₂O(l)

Moles of copper are:

$4.95cm^3\frac{8.95g}{1cm^3} \frac{1mol}{63.55g} = 0.697 moles$

Moles of nitric acid are:

$230mL\frac{1.42g}{mL} \frac{68g}{100g} \frac{1mol}{63.01g}=3.52moles$

As 1 mol of Cu reacts with 4 moles of HNO₃:

0.697 mol Cu × (4mol HNO₃ / 1mol Cu) = 2.79 moles of HNO₃ will react. That means Cu is limiting reactant.

Moles of NO₂ produced are:

0.697 mol Cu × (2mol NO₂ / 1mol Cu) = 1.394 moles of NO₂

Using PV = nRT

Where P is pressure (735torr / 760 = 0.967atm); n are moles (1.394mol); R is gas constant (0.082atmL/molK); T is temperature (28.2°C + 273.15 = 301.35K).

Thus, volume is:

V = nRT / P

V = 1.394mol×0.082atmL/molK×301.35K / 0.967atm

V = 35.6L

The volume of nitrogen oxide formed is 35.6L

3 0

3 years ago