Answer:
T2= 7.3°C
Explanation:
To solve this problem we will use Charles law equation i.e,
V1/T1 = V2/T2
Given data
V1 = 269.7 L
T1 = 6.12 °C
V2= 320.4 L
T2=?
Solution:
Now we will put the values in equation
269.7 L / 6.12°C = 320.4 L / T2
T2= 320.4 L × 6.12°C/ 269.7 L
T2= 1960.85 °C. L /269.7 L
T2= 7.3°C
The concentration of the sodium hydroxide will be 0.016 M
<h3>Stoichiometric problem</h3>
First, the concentration of the diluted nitric acid needs to be found.
m1 = 2.0 M, v1 = 100 mL, v2 = 300.0 mL
m2 = 2 x 100/300 = 0.6667 M
The equation of the reaction goes thus:
The mole ratio is 1:1.
Mole of 0.6667 M, 7.05 mL HNO3 = 0.6667 x 0.00705 = 0.0047 mol
Equivalent mole of NaOH = 0.0047 mol
Molarity of 0.0047 mol, 30.0 mL NaOH = 0.0047/0.3 = 0.016 M
More on stoichiometric problems can be found here: brainly.com/question/27287858
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A calorimeter contains 500 g of water at 25°C.....
the temperature of the water inside the calorimeter is 39.4°C.....
The specific heat of water is 4.18 J/g-°C.
energy needed to heat the water = specific heat * mass * temp difference
= 4.18 J/g-°C * 500 g * (39.4°C - 25°C)
= 4.18*500*14.4
= 30096J
or approx. 30kJ
Answer:
11.3 g.
Explanation:
Hello there!
In this case, since the combustion of butane is:
Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:
Therefore, the resulting mass of water is:
Best regards!