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4 years ago

10

Consider the following balanced equation:

1 answer:

Alenkasestr [34]4 years ago

3 0

Answer:

D

Explanation:

bs I just done that on my work and asked my teacher if it was right she said it was

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Does a low or a high pressure air mass have a greater atmospheric pressure than the surrounding air masses?

Ghella [55]

Answer:

Hello fellow brainlian! here is the answer that you seek:High pressure areas are usually caused by air masses being cooled, either from below (for instance, the subtropical high pressure zones that form over relatively cool ocean waters to the west of Califormia, Africa, and South America

Explanation:

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3 years ago

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When do plants use the process of cellular respiration

d1i1m1o1n [39]

Answer:

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Explanation:

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4 years ago

Mg(s) + HCl (l) --> H2 (g) +MgCl2 (s) A student conducted an experiment with 2g of magnesium and 3 mL of 6M hydrochloric acid

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3 years ago

Ideal He gas expanded at constant pressure of 3 atm until its volume was increased from 9 liters to 15 liters. During this proce

kkurt [141]

Explanation:

The given data is as follows.

P = 3 atm

= $3 atm \times \frac{1.01325 \times 10^{5} Pa}{1 atm}$

= $3.03975 \times 10^{5} Pa$

$V_{1}$ = 9 L = $9 \times 10^{-3} m^{3}$ (as 1 L = 0.001 $m^{3}$ ),

$V_{2}$ = 15 L = $15 \times 10^{-3} m^{3}$

Heat energy = 800 J

As relation between work, pressure and change in volume is as follows.

W = $P \times \Delta V$

or, W = $P \times (V_{2} - V_{1})$

Therefore, putting the given values into the above formula as follows.

W = $P \times (V_{2} - V_{1})$

= $3.03975 \times 10^{5} Pa \times (15 \times 10^{-3} m^{3} - 9 \times 10^{-3} m^{3})$

= 1823.85 Nm

or, = 1823.85 J

As internal energy of the gas $\Delta E$ is as follows.

$\Delta E$ = Q - W

= 800 J - 1823.85 J

= -1023.85 J

Thus, we can conclude that the internal energy change of the given gas is -1023.85 J.

8 0

3 years ago