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4 years ago

How many grams of oxygen gas are contained in a 15 L sample at 1.02 atm and 28°C? Please show your work.

Chemistry

1 answer:

vampirchik [111]4 years ago

4 0

Answer:

0.019 g.

Explanation:

Firstly, we need to find the no. of moles of oxygen gas:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 1.02 atm).

V is the volume of the gas in L (V = 15.0 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 28°C + 273 = 301.0 K).

∴ n = PV/RT = (1.02 atm)(15.0 L)/(0.0821 L.atm/mol.K)(301.0 K) = 0.62 mol.

To find the mass of oxygen gas, we have:

no. of moles = mass/molar mass.

∴ mass of oxygen = (no. of moles)(molar mass) = (0.62 mol)(32.0 g/mol) = 0.019 g.

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kari74 [83]

C. Beryllium. Which has 112 pm

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4 years ago

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ANEK [815]

Answer:

Exit velocity of air is 96.43 m/s.

Explanation:

Given that

$V_1=400\ m/s$

$T_1=25C$

$T_2=100C$

For air

$C_p=1.005\ KJ/kg.K$

Now from first law of thermodynamics for open system at steady state

$h_1+\dfrac{V_1^2}{2000}+Q=h_2+\dfrac{V_2^2}{2000}+w$

For diffuser

$h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}$

We know that

$h=C_pT$

$h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2000}$

$1.005\times 25+\dfrac{400^2}{2000}=1.005\times 100+\dfrac{V_2^2}{2}$

$V_2=96.43\ m/s$

So the exit velocity of air is 96.43 m/s.

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4 years ago

Which compounds are lewis bases? select all that apply. h―c≡c―h h− nh3 ch3ch2ch3?

Alisiya [41]

The Lewis bases are those which can donate electron pair to an acceptor. These electrons can be a pair of electron that is a lone pair, a negative charge or electrons in a $\pi$ -bond.

In $H-C\equiv C-H$ , the presence of $\pi$ -bond make it a Lewis base.
In $H^{-}$ , the presence of negative charge makes it a Lewis base.
In $NH_3$ , the presence of lone pair on nitrogen makes it a Lewis base.
In $CH_3CH_2CH_3$ , there is no lone pair, no $\pi$ -bond, no negative charge so it is not a Lewis base.