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n200080 [17]
3 years ago
9

Three players (A, B and C) shoot three arrows at a target

Chemistry
1 answer:
UkoKoshka [18]3 years ago
6 0

Answer:

Player B

Explanation:

I just did it and I got it right :)

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A container holds 35.8 moles of gas under 10.0 atm of pressure at 70.0 C. What is the volume of the container? 20.6 L 101 L 2080
Pavlova-9 [17]

PV = n RT
P: pressure =10atm
V volume
n number of mole = 35.8 moles
R universal gas constant = 0.082
T: The temperature= 70°C= 343.15 Kelvin

 
V= (n RT) / P = 35.8 x 0.082 x 343.15   / 10 = 100.7 ≈ 101 L

 V = 101L

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3 years ago
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How many moles of water will be generated during the combustion of 0.38 moles of methyl alcohol (CH3OH)? 2CH3OH + 3O2 2CO2 + 4H2
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3 years ago
1 times 2900 inches for a car
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3 years ago
When a 2.00 g sample of KCl is dissolved in water in a calorimeter that has a total heat capacity of 1.28 kJ ⋅ K − 1 , the tempe
Alchen [17]

Answer : The molar heat of solution of KCl is, 17.19 kJ/mol

Explanation :

First we have to calculate the heat of solution.

q=c\times (\Delta T)

where,

q = heat produced = ?

c = specific heat capacity of water = 1.28kJ/K

\Delta T = change in temperature = 0.360 K

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

q=1.28kJ/K\times 0.360K

q=0.4608kJ=460.8J

Now we have to calculate the molar heat solution of KCl.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 460.8 J

m = mass of KCl = 2.00 g

Molar mass of KCl = 74.55 g/mol

\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{2.00g}{74.55g/mole}=0.0268mole

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

\Delta H=\frac{460.8J}{0.0268mole}

\Delta H=17194.029J/mol=17.19kJ/mol

Therefore, the molar heat of solution of KCl is, 17.19 kJ/mol

7 0
3 years ago
I have a question about ionization energy.
ValentinkaMS [17]
Ionization energy is the measure of the extend to which the nucleus attracts the outermost electron
if ionization energy us high than force of attraction Is high so it is not easy to remove and vice versa .
hope you understand.....
5 0
3 years ago
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