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nata0808 [166]
3 years ago
5

How is chemical energy useful? Give three examples.

Chemistry
1 answer:
Leona [35]3 years ago
7 0
1) For energy
2) for chemical energy
3) for very chemically-y-y energy-gy-gy
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An experiment shows that a 250 ?ml gas sample has a mass of 0.430 g at a pressure of 736 mmhg and a temperature of 28 ?c.
dmitriy555 [2]
What we're looking for here is the gas sample's molar mass given its mass, pressure, volume, and temperature. Recalling the gas law, we have

PV = nRT or
n = \frac{PV}{RT}

where R is <span>0.08206 L atm / mol K, P is the given pressure, T is the temperature, and V is the volume.

Before applying the values given, it is important to make sure that they are to be converted to have consistent units with that of R. 
</span>
Thus, we have

P = 736/ 729 = 0.968 atm
T = 28 + 273.15 = 301.15 K
V = 250/1000 = 0.250 L

Now, applying these converted values into the gas law, we have

n = \frac{(0.968 atm)(0.250 L)}{(0.08206 L.atm/mol.K)(301.15 K)}
n = 0.00979 moles

Given that the mass of the sample is 0.430 g, we have

molar mass = \frac{mass}{number of moles} 
molar mass = \frac{0.430}{0.00979} = 43.9

Thus, the gas sample has a molar mass of 43.9 g/mol.
4 0
3 years ago
What is the product(s) of the reaction below?
Lesechka [4]

Answer:

is b

Explanation:

es la b porque estpy estudiando lo mismo

7 0
2 years ago
In Mendelian genetics, when a tall plant is crossed with a short plant, the offspring will be:
Annette [7]
Some offspring may be tall, some may be short, and some may be medium medium sized.
3 0
3 years ago
Read 2 more answers
What term is best described as the rate of a chemical reaction at any instant in time?
Fantom [35]

Answer:

instantaneous rate would be the term.

5 0
2 years ago
At constant pressure, which of these systems do work on the surroundings? A ( s ) + B ( s ) ⟶ C ( g ) A(s)+B(s)⟶C(g) 2 A ( g ) +
Tju [1.3M]

Correct question:

At constant pressure, which of these systems do work on the surroundings?

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

(c) A ( g ) + B ( g ) ⟶ C ( g )

(d) 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

Answer:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

Explanation:

Work done by a system on the surroundings at a constant pressure is given as;

W = -PΔV

Where;

ΔV is gas expansion, that is final volume of the gas minus initial volume of the gas must be greater than zero.

Part (a)

A ( s ) + B ( s ) ⟶ C ( g )

ΔV = 1 - (0) = 1 (expansion)

Part (b)

2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

ΔV = 5 - ( 2+ 2) = 1 (expansion)

Part (c)

A ( g ) + B ( g ) ⟶ C ( g )

ΔV = 1 - ( 1 + 1) = -1 (compression)

Part (d)

2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

ΔV = 3 - ( 4) = -1 (compression)

Thus, systems where there is gas expansion are in part (a) and part (b). The correct answers are:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

4 0
2 years ago
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