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Nadya [2.5K]
3 years ago
9

How many cubic feet of air at standard conditions 1.00 atm are required to inflate a bicycle tote of .50 cu. Ft. To a pressure o

f 3.00 atmospheres?
Chemistry
1 answer:
Alexeev081 [22]3 years ago
3 0
As I am reading the problem, I see they gave you two pressures, one volume and they are asking for another volume. this should give you a hint that you need to use the following formula. 

P1V1= P2V2

P1= 1.00 atm
V1= 0.50 ft³
P2= 3.00 atm
V2= ?

Now we plug the values

(1.00 x 0.50)= (3.00 x V2)

V2= 0.17 ft³
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A researcher wants to test the solubility (property of being dissolved) of salt in water as the temperature of the water increas
Leya [2.2K]
The researcher may first weight the beaker with water and then start to heat the water to a constant temperature, for example 30 °C and then start adding salt and stirring. He should add salt slowly until solid salt starts to become visible and the solution starts becoming cloudy. When this happens, he should quickly weigh the beaker. The increase in mass is the mass of salt dissolved at that temperature.
The procedure is then repeated but at an increased temperature until 5-6 temperatures have been tested.
3 0
3 years ago
How many grams of Co3+ are present in 2.34 grams of cobalt(III) nitrite?
Assoli18 [71]

Answer:

m_{Co^{3+}}=0.563gCo^{3+}

Explanation:

Hello there!

In this case, since these mole-mass relationships are understood in terms of the moles of the atoms forming the considered compound, we first realize that the chemical formula of the cobalt (III) nitrate is Co(NO₃)₃ whereas there is a 1:1 mole ratio of the cobalt (III) ion (molar mass = 58.93 g/mol) to the entire compound. In such a way, we first compute the moles of the salt (molar mass = 58.93 g/mol) and then apply the aforementioned mole ratio to obtain the grams of the required cation:

m_{Co^{3+}}=2.34gCo(NO_3)_3*\frac{1molCo(NO_3)_3}{244.95 gCo(NO_3)_3} *\frac{1molCo^{3+}}{1molCo(NO_3)_3} *\frac{58.93gCo^{3+}}{1molCo^{3+}} \\\\m_{Co^{3+}}=0.563gCo^{3+}

Best regards!

4 0
2 years ago
Describe the relationship between hydrogen oxygen and water
True [87]
Hydrogen is composed of H atom and oxygen is composed of O atom. For water, it is composed by both H and O atom. If you burn hydrogen in oxygen, you can get water. And if you electrolysis water, you can get hydrogen and oxygen.
6 0
3 years ago
Calculate the equilibrium constant for the reaction using the balanced chemical equation and the concentrations of the substance
Oliga [24]

Answer:

4.75 is the equilibrium constant for the reaction.

Explanation:

CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

Equilibrium concentration of reactants :

[CO]=0.0590 M,[H_2O]=0.00600 M

Equilibrium concentration of products:

[CO_2]=0.0410 M,[H_2]=0.0410 M

The expression of an equilibrium constant is given by :

K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}

K_c=\frac{0.0410 M\times 0.0410 M}{0.0590M\times 0.00600 M}

K_c=4.75

4.75 is the equilibrium constant for the reaction.

3 0
3 years ago
A quantity of water is heated from 25.0°C to 36.4°C by absorbing 325 J of heat energy. If the specific heat of water is 4.18 J /
Arlecino [84]

Answer:

6,8 g

Explanation:

c = 4.18 J/(g * °C) = 4180 J / (kg * °C)

t_{1} = 25 °C

t_{2} = 36,4 °C

Q = 325 J

The formula is: Q = c * m * (t_{2} - t_{1})

m = \frac{Q}{c * (t_{2} - t_{1} )}

Calculating:

m = 325 / 4180 * (36,4 - 25) ≈ 0,0068 kg = 6,8 g

6 0
3 years ago
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