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Elis [28]
3 years ago
7

It is 12:00 and you just heard that local officials are tracking a violent storm heading westward, directly for your town and sc

hool. The storm is traveling at 25 km/hr and it is currently 75 kilometers east of your school. If the school day is not shortened, and you remain until two o’clock (the official end of the day), will you and all others in the school get home before the storm hits?
Chemistry
2 answers:
Softa [21]3 years ago
6 0

Answer:

Yes.

Explanation:

i got it correct :)

strojnjashka [21]3 years ago
3 0

Answer:

The answer is : Yes

Explanation:

because it 12:00 and the storm is moving at 25km/hr and it 75 km away that means in three hours the storm would hit your town so at 3:00 the storm would hit at your at ends at 2:00 so yes you would get home before he storm hit.

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The addition of 250.0 J to 30.0 g of copper initially at 22.0°C will change its temperature to what final value? (Specific heat
WINSTONCH [101]

Answer:

Final temperature = 43.53^{\circ} C

Explanation:

Given that,

Heat added, Q = 250 J

Mass, m = 30 g

Initial temperature, T₁ = 22°C

The Specific heat of Cu= 0.387 J/g °C

We know that, heat added due to the change in temperature is given by :

Q=mc\Delta T\\\\Q=mc(T_2-T_1)\\\\\dfrac{Q}{mc}=(T_2-T_1)\\\\T_2=\dfrac{Q}{mc}+T_1

Put all the values,

T_2=\dfrac{250}{30\times 0.387}+22\\\\=43.53^{\circ} C

So, the final temperature is equal to 43.53^{\circ} C.

8 0
3 years ago
How do the rates of the forward and reverse reactions change throughout a reaction?
fenix001 [56]
The rates of the forward and reverse reaction depends on the temperature on which the reaction will proceed, either endothermic of exothermic. it also depends of the concentration of the reactants and products. if the reaction is exothermic, so if the reaction temperature is increased then it will favor the forward reaction, then if the reaction is lowered then it will favor the reverse reaction
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2 years ago
The man responsible for discovering that the atom is mostly space is:
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Rutherford is the man responsible :)

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3 years ago
When 0.0901 mol of an unknown hydrocarbon is burned in a bomb calorimeter, the calorimeter increases in temperature by 2.19°C. I
Alexus [3.1K]

Answer:

The heat of combustion for the unknown hydrocarbon is -29.87 kJ/mol

Explanation:

Heat capacity of the bomb calorimeter = C = 1.229 kJ/°C

Change in temperature of the bomb calorimeter = ΔT = 2.19°C

Heat absorbed by bomb calorimeter = Q

Q=C\times \Delta T

Q=1.229 kJ/^oC\times 2.19^oC=2,692 kJ

Moles of hydrocarbon burned in calorimeter = 0.0901 mol

Heat released on combustion = Q' = -Q = -2,692 kJ

The heat of combustion for the unknown hydrocarbon :

\frac{Q'}{0.090 mol}=\frac{-2,692 kJ}{0.0901 mol}=-29.87 kJ/mol

6 0
3 years ago
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Answer: -5

Explanation:

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