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3 years ago

7

I have doubts about the air pollution puzzle. Can anyone help? Sarah replied, but I can't see her response

1 answer:

Eva8 [605]3 years ago

7 0

Answer:

wdym exactly?

Explanation:

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Zoom in if need to<br> PLEASE HELP DUE TOMARROW

padilas [110]

The best way to do this is to google search each question, or look it up on quizlet.com

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3 years ago

Explain how temperature and precipitation are affected by latitude and the sun's rays.

Contact [7]

Latitude and Temperature

At the equator, the Sun's rays are most direct. This is where temperatures are highest. At higher latitudes, the Sun's rays are less direct. The farther an area is from the equator, the lower its temperature.

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3 years ago

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Two systems with heat capacities 19.9 J mol-1 K-1 and 28.2 ] mol 1 K-1 respectively interact thermally and come to an equilibriu

MAVERICK [17]

Answer : The initial temperature of system 2 is, $19.415^oC$

Explanation :

In this problem we assumed that the total energy of the combined systems remains constant.

$-q_1=q_2$

$m\times c_1\times (T_f-T_1)=-m\times c_2\times (T_f-T_2)$

The mass remains same.

where,

$C_1$ = heat capacity of system 1 = 19.9 J/mole.K

$C_2$ = heat capacity of system 2 = 28.2 J/mole.K

$T_f$ = final temperature of system = $30^oC=273+30=303K$

$T_1$ = initial temperature of system 1 = $45^oC=273+45=318K$

$T_2$ = initial temperature of system 2 = ?

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

$-19.9J/mole.K\times (303-318)K=28.2J/mole.K\times (303-T_2)K$

$T_2=292.415K$

$T_2=292.415-273=19.415^oC$

Therefore, the initial temperature of system 2 is, $19.415^oC$

8 0

4 years ago

Predict the products of the thermal decomposition of strontium nitrate and show the reaction as a word equation.

Schach [20]

yeag

Explanation:

2SrO + 4NO2 + O. The thermal decomposition of strontium nitrate to produce strontium oxide, nitrogen dioxide and oxygen. This reaction takes place at a temperature of over 570°C

6 0

3 years ago

Q Q 3. (08.02 MC)

AnnyKZ [126]

Answer: A volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.

Explanation:

Given: $V_{1}$ = ?, $M_{1}$ = 0.55 M

$V_{2}$ = 100.0 mL, $M_{2}$ = 2.50 M

Formula used to calculate the volume of KBr is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\0.55 M \times V_{1} = 2.50 M \times 100.0 mL\\V_{1} = 455 mL$

Thus, we can conclude that a volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.

7 0

3 years ago

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