A bicycle rider is applying a force of 20 N while heading south against a wind blowing from the south with a force of 5 N

This reaction is sometimes used to inflate life rafts, weather balloons, and the like, where a simple, compact means of generati

tia_tia [17]

Answer: 10.2 grams

Explanation:

The balanced chemical reaction is :

$CaH_2(s)+2H_2O(l)\rightarrow Ca(OH)_2(aq)+2H_2(g)$

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 740 torr = 0.97 atm (760torr=1atm)

V= Volume of the gas = 12.0 L

T= Temperature of the gas = 19°C = 292 K $0^0C=273K$

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas

$n=\frac{PV}{RT}=\frac{0.97\times 12.0}{0.0821\times 292}$

$n=0.48$

According to stoichiometry:

2 moles of hydrogen are generated by = 1 mole of $CaH_2$

Thus 0.48 moles of hydrogen are generated by = $\frac{1}{2}\times 0.48=0.24$ moles of $CaH_2$

Mass of $CaH_2=moles\times {\text {Molar mass}}=0.24mol\times 42g/mol=10.2g$

Thus 10.2 grams of $CaH_2$ are needed to generate 12.0 L of hydrogen gas if the pressure of hydrogen is 740. torr at 19°C

5 0

3 years ago

If oxygen at 128 kpa is allowed to expand at constant temp until it's pressure is 101.3 kpa how much larger will the volume beco

LenaWriter [7]

which means that the volume increased by 26.4 mL in order to compensate for the decrease in pressure.

Like I said, depends on what your initial volume was, but that's how you think of it.

Hope this helped!

6 0

2 years ago

What types of material might have survived 85 years of weather? answer?

Mashutka [201]

Hey You!

Metal and Minerals may have survived 85 years of weather.

5 0

3 years ago

What is the density of an object that has a mass of 6.5 g and when placed in water displaces the volume from 4.5mL to 11.8mL? ro

KengaRu [80]

Answer:

$\rho =0.9g/mL$

Explanation:

Hello,

In this case, due to the volume displacement caused the by the object's submersion, it's volume is:

$V=11.8mL-4.5mL=7.3 mL$

In such a way, considering the mathematical definition of density, it turns out:

$\rho =\frac{m}{V}=\frac{6.5g}{7.3mL}\\ \\\rho =0.89g/mL$

Rounding to the nearest tenth we finally obtain:

$\rho =0.9g/mL$

Regards.

3 0

3 years ago

Suppose Gabor, a scuba diver, is at a depth of 15m15m. Assume that: The air pressure in his air tract is the same as the net wat

alexandr1967 [171]

Answer:

The ration of molar concentration is "2.5".

Explanation:

The given values are:

Average density of salt water,

= $1.03 \ g/cm^3$

Net pressure,

= $2.00 \ atm$

Increase in pressure,

= $1.00 \ atm$

Now,

The under water pressure will be:

= $\frac{15 \ m}{10}\times 1 \ atm +1 \ atm$

= $1.5\times 1+1$

= $1.5+1$

= $2.5 \ atm$

hence,

The ratio will be:

= $\frac{(\frac{n}{V})_{15m} }{(\frac{n}{V})_{surface} }$

or,

= $\frac{P}{P_s}$

= $\frac{2.5}{1}$

= $2.5$

7 0

2 years ago