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wolverine [178]
2 years ago
5

Strontium and chlorine combine in only one ratio: one strontium atom for every two chlorine atoms. Based on this

Chemistry
1 answer:
Vinvika [58]2 years ago
4 0

The combinations of Strontium and chlorine  that are possible are only those in which strontium and chlorine combine in the ratio of 1:2.

1) The possible combinations are

  • 4 strontium atoms and 8 chlorine
  • 2.7 * 10^19 strontium atoms and5.4 * 10^19 chlorine atoms

2) The combinations that are not possible are

  • 20 strontium atoms and 60 chlorine atoms
  • 130 billion strontium atoms and 195 billion chlorine

We have to work out the ratio of Strontium and chlorine in each of the given  combinations in the question. Only the combinations in which the ratio of Strontium and chlorine is 1:2 is possible.

First case:

4 strontium atoms and 8 chlorine atoms gives a Strontium and chlorine ratio of 1:2 so it is possible.

Second case:

20 strontium atoms and 60 chlorine atoms gives a Strontium and chlorine ratio of 1:3 hence it is not possible.

Third case:

2.7 * 10^19 strontium atoms and 5.4 * 10^19 chlorine atoms gives a Strontium and chlorine ratio of 1:2 hence it is possible.

Fourth case:

130 billion strontium atoms and 195 billion chlorine atoms gives a Strontium and chlorine ratio of 1:1.5 hence it is not possible.

Learn more: brainly.com/question/9743981

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garri49 [273]

Answer: Francesco Redis Experiment part 1 : The maggots were not observed in setup B and setup C because the jars were covered.

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Pasteur Experiment

part 1:In the pastures Experiment the microorganisms come from the air.

part 2:Louis Pasteur's pasteurization experiment illustrates the fact that the spoilage of liquid was caused by particles in the air rather than the air itself.

Explanation:

8 0
3 years ago
How many kilograms of water must be added to 6.07 grams of oxalic acid (H2C2O4) to make a 0.025 m solution?
uranmaximum [27]
This dilution problem uses the equation
M
a
V
a
=
M
b
V
b

M
a
= 6.77M - the initial molarity (concentration)
V
a
= 15.00 mL - the initial volume
M
b
= 1.50 M - the desired molarity (concentration)
V
b
= (15.00 + x mL) - the volume of the desired solution
(6.77 M) (15.00 mL) = (1.50 M)(15.00 mL + x )
101.55 M mL= 22.5 M mL + 1.50x M
101.55 M mL - 22.5 M mL = 1.50x M
79.05 M mL = 1.50 M
79.05 M mL / 1.50 M = x
52.7 mL = x
59.7 mL needs to be added to the original 15.00 mL solution in order to dilute it from 6.77 M to 1.50 M.
I hope this was helpful.
4 0
2 years ago
In a car engine gasoline is burning to create mechanical energy which of the following statements is true? A. Some energy is los
Anvisha [2.4K]

Answer:

Some energy is lost as heat

Explanation:

It is correct to say that as the gasoline is converted to mechanical energy in the automobile engine, some of the energy is lost as heat.

Heat energy is on of the ways energy is lost in any system. The movement of mechanical parts and even the combustion of the gasoline produces heat energy.

These energy are usually lost to the environment.

4 0
3 years ago
Calculate the equilibrium constant for the decomposition of water 2h2o(l)  2h2(g) + o2(g) at 25°c, given that g°f (h2o(l)) = –
kow [346]

Answer:

2.6 ×10^-42

Explanation:

From

∆G= -RTlnK

∆G= -237.2 KJmol-1 or -237.2×10^3 Jmol-1

R= 8.314 Jmol-1K-1

T= 25°C + 273= 298K

-237.2×10^3= 8.314 × 298 × ln K

ln K= -237.2×10^3/2477.572

K = 2.6 ×10^-42

3 0
3 years ago
How many grams of argon would it take to fill a large light bulb with a volume of 0.745 L at STP?
EastWind [94]

Answer:

Mass = 1.33 g

Explanation:

Given data:

Mass of argon required = ?

Volume of bulb = 0.745 L

Temperature and pressure = standard

Solution:

We will calculate the number of moles of argon first.

Formula:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

By putting values,

1 atm ×0.745 L = n × 0.0821 atm.L/mol.K× 273.15 K

0.745 atm. L = n × 22.43 atm.L/mol

n = 0.745 atm. L / 22.43 atm.L/mol

n = 0.0332 mol

Mass of argon:

Mass = number of moles × molar mass

Mass = 0.0332 mol × 39.95 g/mol

Mass = 1.33 g

8 0
3 years ago
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