Answer:Answer:5.95 LExplanation:V1/V2 = T1/T24.82 L/V2 = 243.1K/300.1KV2 = 4.82*300.1/243.1 L = 5.95 L
Explanation: The volume of the balloon is 400 m3. To what temperature must the air in the balloon be heated before the balloon will lift off. (Air density at 10 oC is 1.25 kg/m3.)
Answer:
14.8 g/mol
Explanation:
Starting on the left, count in 3 numbers (14.8), then round using the next number (0.02).
Because 2 is less than 5, the number stays the same and is not rounded up.
Answer:
Explanation:
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In this case, according to this equilibrium temperature problem, we can set up the following equation to relate the mass, specific heat and temperature change:
Thus, we solve for the mass of cobalt as shown below:
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<span>Warmer air is pushed upward and out of the way by cooler air.</span>
1.
V = 200 mL (volume)
c = 3 M = 3 mol/L (concentration)
First we convert mL to L:
200 mL = 0.2 L
Then we calculate the moles using the formula: n = V × c = 0.2 L × 3 mol = 0.6 mol
Finally, we just use the molar mass of CaF2 to calculate the actual mass:
molar mass = 78 g/mol
The formula is: m = n × mm (mass = moles × molar mass)
m = 0.6 mol × 78 g/mol = 46.8 g
2.
For this question the steps are exactly like the first question.
V = 50mL = 0.05 L
c = 12 M = 12 mol/L
n = V × c = 0.05 L × 12 mol/L = 0.6 mol
molar mass (HCl) = 36.5 g/mol
m = n × mm = 0.6 mol × 36.5 g/mol = 21.9 g.
3.
The steps for this question are the opposite way.
m(K2CO3) = 250 g
molar mass = 138 g/mol
n = m ÷ mm = 1.81 mol
c = 2 mol/L
V = n ÷ c = 1.81 mol ÷ 2 mol/L = 0.905 L = 905 mL
