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3 years ago

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Two rocks one weighing 100m and the other weiging 200n are drpped from a 50-m cliff at the same time when both rocks are 10m fro

m the ground and air friction is ignored which is the same

1 answer:

Ratling [72]3 years ago

8 0

Answer:

Ignore all friction

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555mL of gas is combined at standard temperature and pressure(STP). The volume changes to 660.mL at 30.5 °C. What is the new pre

liberstina [14]

Answer:0.94

Explanation:

4 0

3 years ago

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

bija089 [108]

<u>Answer:</u> The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u> $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

<u>Equation 2:</u> $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$

We are given:

$K_1=K_b$

$K_2=\frac{1}{K_w}$

Putting values in above equation, we get:

$K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}$

Hence, the value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

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3 years ago

Determine the formula weights of each of the following compounds.Part A) Nitrous oxide, N2O, known as laughing gas and used as a

densk [106]

Answer:

See explanation

Explanation:

a) Nitrous oxide (N2O) has a molar mass of 44.014 amu. It has 2 nitrogen atoms each with a mass of 14.007 amu and 1 oxygen atom with a mass of 16.0 amu.

Percentage nitrogen = (2*14.007 amu/ 44.014 amu ) *100% = 63.6%

Percentage oxygen = (16 amu/44.014 amu) * 100% = 36.4 %

63.6% + 36.4% = 100%

b) Benzoic acid (C7H6O2) has a molar mass of 122.13 amu. It has 6 hydrogen atoms each with a mass of 1.01 amu; it has 7 carbon atoms each with a mass of 12.01amu and 2 oxygen atoms with a mass of 16.0 amu.

Percentage hydrogen = (6*1.01 amu / 122.13 amu)*100% = 4.96%

Percentage carbon = (7*12.01 amu/ 122.13 amu)*100% = 68.8%

Percentage oxygen = (2*16 amu/ 122.13 amu) *100% = 26.2%

c) Magnesium hydroxide (Mg(OH)2) has a molecular mass of 58.32 amu. It has 2 hydrogen atoms each with a mass of 1.01 amu; it has 1 magnesium atom with a mass of 24.3 amu and two oxygen atoms with a mass of 16.0 amu.

Percentage hydrogen = (2*1.01 amu/ 58.32 amu) *100% = 3.46 %

Percentage magnesium = (24.3 amu/58.32 amu)*100% = 41.7%

Percentage oxygen = (2*16 amu/58.32 amu)*100% = 54.9%

d) Urea CO(NH2)2 has a molecular mass of 60.064 amu. It has 2 Nitrogen atoms each with a mass of 14.007 amu, 4 hydrogen atoms each with a mass of 1.01 amu,1 carbon atom with a mass of 12.01 amu and 1 oxygen atom with a mass of 16.0 amu.

Percentage nitrogen = (2*14.007 amu/ 60.064amu)*100% = 46.6%

Percentage hydrogen = (4*1.01 amu/60.064amu)*100% = 6.72%

Percentage carbon = (12.01 amu/60.064amu)*100% = 20.0%

Percentage oxygen = (16 amu/60.064amu)*100% = 26.6%

e) Osopentyl acetate (C7H14O2) has a molecular mass of 130.2 amu. It has 14 hydrogen atoms each with a mass of 1.01 amu,7 carbon atoms each with a mass of 12.01 amu and 2 oxygen atom with a mass of 16.0 amu.

Percentage hydrogen = (14*1.01 amu/130.2 amu)*100% = 10.8%

Percentage carbon = (7*12.01 amu/130.2 amu)*100% = 64.6%

Percentage oxygen =(2*16 amu/130.2 amu)*100% = 24.6%

7 0

3 years ago

IBNEED HELP ASAP<br> NO BULL I NEED ACTUAL HELP

Alja [10]

Answer:

god says dont use any assault language okay?

Explanation:

5 0

3 years ago

What is the pressure, in atmospheres, of a 0.108-mol sample of helium gas at a temperature of 20.0°C if its volume is 0.505 L?

SpyIntel [72]

Answer:

P = 5.14ATM

Explanation:

Number of moles = 0.108moles

Temperature (T) = 20°C = 20 + 273.15 = 293.15K

Volume V = 0.505L.

Pressure (P) = ?

R = 0.082J/mol.K

From ideal gas equation,

PV = nRT

P = nRT / V

P = (0.108 * 0.082 * 293.15) / 0.505

P = 2.596 / 0.505

P = 5.14ATM

The pressure of the gas is 5.14ATM

5 0

3 years ago