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xxMikexx [17]
3 years ago
7

36 g of water vapor takes up how many liters at room temperature and pressure of 293K and 100 K PA

Chemistry
2 answers:
qaws [65]3 years ago
7 0

Answer:-

48.72 L

Explanation:-

Chemical formula of water= H2O

Mass of water = 36 g

Molar mass of water = 1 x 2 + 16 x 1 = 18 g mol-1

Number of moles of water = n = 36 g / (18 g mol-1)

= 2 moles.

Temperature T = 293 K

Pressure = P = 100 K Pa

Universal gas constant R = 8.314 L KPa K -1 mol-1

Using the relation PV = nRT

we get Volume V = nRT/P

Plugging in the values

V = (2 moles ) x ( 8.314 L KPa K -1 mol-1) x ( 293 K) / 100 K Pa

= 48.72 L

crimeas [40]3 years ago
6 0

100kpa is correct .100 kpa is correct

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From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp
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<u>Answer:</u> The concentration of NH_3 required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of NiC_2O_4, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of NiC_2O_4 = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M

For the given chemical equations:

NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}

Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9

Net equation: NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?

To calculate the equilibrium constant, K for above equation, we get:

K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48

The expression for equilibrium constant of above equation is:

K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}

As, NiC_2O_4 is a solid, so its activity is taken as 1 and so for C_2O_4^{2-}

We are given:

[[Ni(NH_3)_6]^{2+}]=0.016M

Putting values in above equations, we get:

0.48=\frac{0.016}{[NH_3]^6}}

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7 0
3 years ago
John dissolves .5g of a white powder in 25g of benzene (FP 5oC) (kf benzene is 5.1) and finds the solution freezes at 3.7oC. Det
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Answer:

The compound has a molar mass of 78.4 g/mol

Explanation:

Step 1: data given

Mass of a sample = 0.5 grams

Mass of benzene = 25 grams

Freezing poing = 5 °C

Kf of benzene = 5.1 °C/m

Freezing point solution = 3.7 °C

Step 2: Calculate molality

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 5.0 - 3.7 = 1.3 °C

⇒with i = the can't hoff factor = 1

⇒with Kf = the freezing point depression constant of benzene = 5.1 °C/m

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m = 1.3 / 5.1

m = 0.255 moles /kg

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3 years ago
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Answers:

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              (c)  204.44 °C and 477.44 K

              (d)  -320.8 °F and -196 °C

Explanation:

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Converting °F into °C;

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Converting °C into K;

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Converting K into °C;

                                   °C  =  K - 273

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