6. The diagram below shows a simplified floor plan for a small house. As the day goes

A certain reaction has an activation energy of 28.90 kJ / mol. At what Kelvin temperature will the reaction proceed 5.00 times f

BabaBlast [244]

Answer: 365 K

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $T_1$ = 1.00

$K_2$ = rate constant at $T_2$ = 5.00

$Ea$ = activation energy for the reaction = 28.90 kJ/mol= 28900 j/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 313 K

$T_2$ = final temperature = ?

Now put all the given values in this formula, we get

$\log (\frac{5.00}{1.00})=\frac{28900}{2.303\times 8.314J/mole.K}[\frac{1}{313K}-\frac{1}{T_2K}]$

$0.69=\frac{28900}{2.303\times 8.314J/mole.K}[\frac{1}{313K}-\frac{1}{T_2K}]$

$T_2=365K$

Therefore, 365 K is required to increase the reaction rate by 5.00 times.

4 0

3 years ago

Solutions of mercury (I) nitrate and potassium bromide are mixed

Afina-wow [57]

With that informatio you can:

1) Write the chemical equation
2) Balance the chemical equation
3) State the molar ratios
4) Predict if precipitation occurs.

I will do all four, for you:

1) Chemical equation:

mercury(I) nitrate potassium bromide mercury(I) bromide potassium nitrate
Hg2(NO3)2 + KBr → Hg2Br2 + KNO3

2) Balanced chemical equation

Hg2(NO3)2 + 2KBr → Hg2Br2 + 2KNO3

3) Molar ratios or proportions:

1 mol Hg2(NO3)2 : 2 mol KBr : 1 mol Hg2Br2 : 2 mol KNO3

4) Prediction of precipitation.

You can use the solubility rules or a table of solubilities. I found in a table of solutiblities that mercury(I) bromide is insoluble and potassium bromide is soluble, Then you can predict that the precipitation of mercury(I) bromide will occur.

7 0

3 years ago

1. How many joules must be added to 10.0 g of water to raise its temperature from 10°C to 15°C?

weqwewe [10]

Answer:

209.3 Joules require to raise the temperature from 10 °C to 15 °C.

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m × c × ΔT

Given data:

mass of water = 10 g

initial temperature T1= 10 °C

final temperature T2= 15 °C

temperature change =ΔT= T2-T1 = 15°C - 10°C = 5 °C

Energy or joules added to increase the temperature Q = ?

Solution:

We know that specific heat of water is 4.186 J/g .°C

Q = m × c × ΔT

Q = 10 g × 4.186 J/g .°C × 5 °C

Q = 209.3 J

6 0

3 years ago

A gas balloon has a volume of 80.0 mL at 300K, and a pressure of 50.0 kPa. If the pressure changes to 80 kPa and the temperature

fomenos

Answer: The new volume is 53.3 ml

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 50.0 kPa

$P_2$ = final pressure of gas = 80.0 kPa

$V_1$ = initial volume of gas = 80.0 ml

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $300K$

$T_2$ = final temperature of gas = $320K$

Now put all the given values in the above equation, we get:

$\frac{50.0\times 80.0}{300}=\frac{80.0\times V_2}{320}$

$V_2=53.3ml$

The new volume is 53.3 ml

4 0

2 years ago

Passenger airplanes take off in the troposphere but can eventually move into

zhuklara [117]

Well that is because the stratosphere has lower temperatures and low air density. This gives the airplane less air pressure and with low air density they can fly faster :) Hope this helps.

5 0

2 years ago