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Lubov Fominskaja [6]
2 years ago
9

A family of four lives in a three-bedroom house and uses an average of 900 kWh of electricity per month. The family cools their

house for three months during the summer with two window-unit air conditioners that each use 350 kWh of electricity per month. Which of the following is the percentage of the family’s total annual electricity that is used to run the two air conditioners for the three summer months?
9,7%

19.4%

38.8%

77.8%
Chemistry
1 answer:
agasfer [191]2 years ago
5 0
<h2>Answer:</h2><h2>The percentage of the family’s total annual electricity that is used to run the two air conditioners for the three summer months = 19.4 %</h2>

Explanation:

Average electricity consumed per month = 900 kWh

The family cools their house for three months during the summer with two window-unit air conditioners

The power consumed by one window-unit air conditioners = 350 kWh

The power consumed by two window-unit air conditioners = 350(2) = 700 kWh

Power consumed for two air conditioners for the three summer months = 700 (3) = 2100 kWh

Total power consumed for 1 year = 900 (12) = 10800kWh

The percentage of the family’s total annual electricity that is used to run the two air conditioners for the three summer months = \frac{2100}{10800}100 = 19.4 %

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mixer [17]

Answer:

Number of moles nitric acid in the cylinder is 400.539g/mol.

Explanation:

From the given,

Weight of empty gas cylinder W_{1}= 90.0 lb= 40823.3 gramsWeight of full cylinder[tex]W_{2} = 116.5 lb= 52843.511 gramsThe critical temperature = 287 KThe critical pressure 54.3 LMolar mass of nitric acid = [tex]M_{NO} = 30.01 g/mol

Number of moles nitric acid = n_{NO} =?

The mass of nitric acid in the cylinder = W_{NO}=W_{2}-W_{1}

=52843.511-40823.3 =12,020.2g

Number of moles of nitric acid =

\frac{Given\,mass}{Molar\,mass}=\frac{12,020.2}{30}=400.539g/mol

Therefore, number of moles nitric acid in the cylinder is 400.539g/mol.

3 0
2 years ago
Aluminum–lithium (Al–Li) alloys have been developed by the aircraft industry to reduce the weight and improve the performance of
gtnhenbr [62]

Answer:

The concentration of Li (in wt%) is 3,47g/mol

Explanation:

To obtain the 2,42g/cm³ of density:

2,42g/cm³ = 2,71g/cm³X + 0,534g/cm³Y <em>(1)</em>

<em>Where X is molar fraction of Al and Y is molar fraction of Li.</em>

X + Y = 1 <em>(2)</em>

Replacing (2) in (1):

Y = 0,13

Thus, X = 0,87

The weight of Al and Li is:

0,87*26,98g/mol = 23,4726 g of aluminium

0,13*6,941g/mol = 0,84383 g of lithium

The concentration of Li (in wt%) is:

0,84383g/(0,84383g+23,4726g) ×100= <em>3,47%</em>

6 0
1 year ago
A 2.50 g sample of zinc is heated, and then placed in a calorimeter containing 65.0 g of water. Temperature of water increases f
swat32

718.65 degrees is the initial temperature of the zinc metal sample.

Explanation:

Data given:

mass of zinc sample = 2.50 grams

mass of water  = 65 grams

initial temperature of water = 20 degrees

final temperature of water = 22.5 degrees

ΔT  = change in temperature of water is 2.50 degrees

specific heat capacity of zinc cp= 0.390 J/g°C

initial temperature of zinc sample = ?

cp of water = 4.186 J/g°C

heat absorbed = heat released (no heat loss)

formula used is

q = mcΔT

q water = 65 x 4.286 x 2.5

q water = 696.15 J

q zinc = 2.50 x 0.390 x (22.50- Ti)

equating the two equations

696.15 = - 22.50+ Ti

Ti = 718.65 degrees is the initial temperature of zinc.

6 0
2 years ago
Why are the electrons in a nitrogen-phosphorus covalent bond not shared equally? which atom do the electrons spend more time aro
a_sh-v [17]
<span>Electrons in a nitrogen-phosphorus covalent bond are not shared equally because nitrogen and phosphorus do not have the same electronegativity. The atoms spend more time around the most electronegative atom nitrogen.</span>
6 0
2 years ago
The compound 1,1-difluoroethane decomposes at elevated temperatures to give fluoroethylene and hydrogen fluoride: CH3CHF2(g) → C
Maru [420]

Answer : The final temperature would be, 791.1 K

Explanation :

According to the Arrhenius equation,

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

or,

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where,

K_1 = rate constant at 460^oC = 5.8\times 10^{-6}s^{-1}

K_2 = rate constant at T_2 = 4\times K_1

Ea = activation energy for the reaction = 265 kJ/mol = 265000 J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 460^oC=273+460=733K

T_2 = final temperature = ?

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

\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]

T_2=791.1K

Therefore, the final temperature would be, 791.1 K

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