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3 years ago

Pls help. I have the question in a ss

Chemistry

1 answer:

gladu [14]3 years ago

4 0

Answer:

Explanation:

The hypothesis was that speed would increase with the addition of mass from passengers. The data shows that when passengers/mass are added, speed decreases.

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A 0.100 mile sample of gas is at a temperature of 85.0 degrees C and a volume of 3.47 L. What is the pressure of the gas( in mm

vladimir1956 [14]

Answer: 64.6 mmHg

Explanation:

Given that:

Volume of gas V = 3.47L

(since 1 liter = 1dm3

3.47L = 3.47dm3)

Temperature T = 85.0°C

Convert Celsius to Kelvin

(85.0°C + 273 = 358K)

Pressure P = ?

Number of moles of gas N = 0.100 mole

Note that Molar gas constant R is a constant with a value of 0.0082 ATM dm3 K-1 mol-1

Then, apply ideal gas equation

pV = nRT

p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)

p x 3.47dm3 = 0.29 atm dm3

p = (0.29 atm dm3 / 3.47 dm3)

p = 0.085 atm

Recall that pressure of the gas is required in mm hg, so convert 0.085 atm to mm Hg

If 1 atm = 760 mm Hg

0.085atm = 0.085 x 760

= 64.6 mm Hg

Thus, the pressure of the gas is 64.6 mm hg

6 0

3 years ago

what does it mean when the h3o+ to x- distance doesn't change much as the first 4 to 5 water molecules are added

podryga [215]

If X is an equivalent base to H₂O
HX is an equivalent acid to H₃O⁺
HX is a stronger acid than H₃O⁺
HX is not an acid
X⁻ is a stronger base than H₂O
HX is a weaker acid than H₃O⁺
X⁻ is a weaker base than H₂O
X⁻ is not a base.

The correct response or this is
X⁻ is a stronger base than H₂O
HX is a weaker acid than H₃O⁺

7 0

3 years ago

Read 2 more answers

On a hiking trip, you see a large tree root has grown into the cracks of a rock and caused it to break apart. This is an example

Vadim26 [7]

Hydrologic Weathering I think

3 0

3 years ago

How much heat is released when you condense 93.9 g of water vapor? Show work please ❤️

sammy [17]

Answer:

211.9 J

Explanation:

The molecules of water release heat during the transition of water vapor to liquid water, but the temperature of the water does not change with it.

The amount of heat released can be represented by the formula:

$Q=mL_e$

where $Q$ = heat energy, $m$ = mass of water and $L$ = latent heat of evaporation.

The latent heat of evaporation for water is $L=2257 kJ/kg$ and the mass of the water is $m=93.9 g=0.0939 kg$ .

The amount of heat released in this process is:

$Q=mL_e = (0.0939kg)(2257 kJ/kg)=$ 211.9 J

7 0

3 years ago

Complete the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Ty

anastassius [24]

Answer :

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression of $HC_2H_3O_2$ will be:

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression of $Co(H_2O)_6^{3+}$ will be:

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression of $CH_3NH_3^+$ will be:

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

3 0

3 years ago