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2 years ago

Las Vegas Raiders' running back Josh Jacobs is 72 inches tall and has mass of 100 kg. Los

Chemistry

1 answer:

gladu [14]2 years ago

4 0

Answer:

Josh is running at a speed of 9.09 yards per second.

Josh's velocity is 9.09 East.

Josh's force is 900 N.

Aaron Donald's force is 845 N.

Yes Josh scores the touchdown because he is faster and has more mass than Donald.

Explanation:

Josh scores the touchdown as he is heavier and faster than Donald.

Formulas are:

Force= mass x acceleration

Speed= distance divided by time.

Velocity= distance divided by time.

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A weather balloon is filled with helium that occupies a volume of 5.00 × 10^4 L at 0.995 atm and 32.0°C. After it is released, i

DaniilM [7]

Answer:

The new volume is 5.913*10^4 L

Explanation:

Step 1: Write out the formula to be used:

Using general gas equation;

P1V1 / T1 =P2V2 /T2

V2 = P1V1T2 / P2T1

Step 2: write out the values given and convert to standard unit's where necessary

P1 = 0.995atm

P2 0.720atm

V1 = 5*10^4 L

T1 = 32°C = 32+ 273 = 305K

T2 = -12°C = -12 + 273 = 261K

Step 3: Equate your values and do the calculation:

V2 = 0.995 * 5*10^4 * 261 / 0.720 * 305

V2 = 1298.475 * 10^4 / 219.6

V2 = 5.913 * 10^4 L

So the new volume of the balloon is 5.913*10^4 L

3 0

3 years ago

A gas mixture contains 3.00 atm of H2 and 1.00 atm of O2 in a 1.00 L vessel at 400K. If the mixture burns to form water while th

sleet_krkn [62]

Answer:

$p_{H_2O}=2.00atm$

Explanation:

Hello!

In this case, according to the following chemical reaction:

$2H_2+O_2\rightarrow 2H_2O$

It means that we need to compute the moles of hydrogen and oxygen that are reacting, via the ideal gas equation as we know the volume, pressure and temperature:

$n_{H_2}=\frac{3.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0914molH_2 \\\\n_{O_2}=\frac{1.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0305molH_2$

Thus, the yielded moles of water are computed by firstly identifying the limiting reactant:

$n_{H_2O}^{by\ H_2} = 0.0914molH_2*\frac{2molH_2O}{2molH_2} =0.0914molH_2O\\\\n_{H_2O}^{by\ O_2} = 0.0305molO_2*\frac{2molH_2O}{1molO_2} =0.0609molH_2O$

Thus, the fewest moles of water are 0.0609 mol so the limiting reactant is oxygen; in such a way, by using the ideal gas equation once again, we compute the pressure of water:

$p_{H_2O}=\frac{0.0609molH_2O*0.08206\frac{atm*L}{mol*K}*400K}{1.00L}\\\\ p_{H_2O}=2.00atm$