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Marat540 [252]
3 years ago
5

A 15.0 ml sample of gas at 10.0 degree Celsius and 760 torr changes to a pressure of 1252 torr at 35.0 degree Celsius. What is t

he volume of the gas?
Chemistry
1 answer:
netineya [11]3 years ago
4 0

Answer:

9.91 mL

Explanation:

Using the combined gas law equation as follows;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (torr)

P2 = final pressure (torr)

V1 = initial volume (mL)

V2 = final volume (mL)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

V1 = 15.0mL

V2 = ?

P1 = 760 torr

P2 = 1252 torr

T1 = 10°C = 10 + 273 = 283K

T2 = 35°C = 35 + 273 = 308K

Using P1V1/T1 = P2V2/T2

760 × 15/283 = 1252 × V2/308

11400/283 = 1252V2/308

Cross multiply

11400 × 308 = 283 × 1252V2

3511200 = 354316V2

V2 = 3511200 ÷ 354316

V2 = 9.91 mL

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Despite not knowing mechanistic details of the adsorption of water onto silica gel, from the information provided, you should be
navik [9.2K]

Answer:

G<0, spontanteous

H<0, from equation

S>0, gas to solid

Explanation:

The small bags of silica gel you often see in a new shoe box are placed there to control humidity. Despite its name, silica gel is a solid. It is a chemically inert, highly porous, amorphous form of SiO2. Water vapor readily adsorbs onto the surface of silica gel, so it acts as a desiccant. Despite not knowing mechanistic details of the adsorption of water onto silica gel, from the information provided, you should be able to make an educated guess about the thermodynamic characteristics of the process. Predict the signs of ΔG, ΔH, and ΔS.

G<0, spontanteous

H<0, from equation

S>0, gas to solid

8 0
3 years ago
10) In order to make spaghetti cook faster, a chef adds salt to water. How many moles of salt would he need to add to 1.0 kg wat
Vadim26 [7]
The answer is 4.9 moles.
Solution: 
Using the equation for boiling point elevation Δt,
     Δt = i Kb m 
we can rearrange the expression to solve for the molality m of the solution:
     m = Δt / i Kb 

Since we know that pure water boils at 100 °C, and the Ebullioscopic constant Kb for water is 0.512 °C·kg/mol, 
     m = (105°C - 100°C) / (2 * 0.512 °C·kg/mol)
         = 4.883 mol/kg 

From the molality m of the solution of salt added in a kilogram of water, we can now find the number of moles of salt: 
     m = number of moles / 1.0kg
     number of moles = m*1.0kg 
                                  = (4.883 mol/kg) * (1.0kg)
                                  = 4.9 moles
4 0
3 years ago
Question
Black_prince [1.1K]

Answer:

efef2w

Explanation:

7 0
2 years ago
Which statement best compares the energy involved in melting with the energy involved in boiling for a given liquid?
Soloha48 [4]
C its c trust me i just took the quiz
4 0
3 years ago
Read 2 more answers
I need help on both a and b of question 1
marishachu [46]

Answer:

(a) -0.00017 M/s;

(b) 0.00034 M/s

Explanation:

(a) Rate of a reaction is defined as change in molarity in a unit time, that is:

r = \frac{\Delta c}{\Delta t}

Given the following reaction:

2 N_2O_5 (g)\rightleftharpoons 4 NO_2 (g) + O_2 (g)

We may write the rate expression in terms of reactants firstly. Since reactants are decreasing in molarity, we're adding a negative sign. Similarly, if we wish to look at the overall reaction rate, we need to divide by stoichiometric coefficients:

r = -\frac{\Delta [N_2O_5]}{2 \Delta t}

Reaction rate is also equal to the rate of formation of products divided by their coefficients:

r = \frac{\Delta [NO_2]}{4 \Delta t} = \frac{\Delta [O_2]}{\Delta t}

Let's find the rate of disappearance of the reactant firstly. This would be found dividing the change in molarity by the change in time:

r_{N_2O_5} = \frac{0.066 M - 0.100 M}{200.00 s - 0.00 s} = -0.00017 M/s

(b) Using the relationship derived previously, we know that:

-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}

Rate of appearance of nitrogen dioxide is given by:

r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t}

Which is obtained from the equation:

-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}

If we multiply both sides by 4, that is:

-\frac{4 \Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{\Delta t}

This yields:

[tex]r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t} = -2\frac{\Delta [N_2O_5]}{ \Delta t} = -2\cdot (-0.00017 M/s) = 0.00034 M/s[tex]

5 0
2 years ago
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