A food provides enough energy to raise the temperature of 2000 grams of water by 10°c contains 20 KCa.
1 Calorie is the amount of heat needed to elevate one gram of water to one degree Celsius temperature at one atmosphere of pressure.
The term "kilocalorie" is used to describe the amount of energy needed to increase 1 L of water's temperature by one degree Celsius at sea level.
Given
Mass of water (m) = 2000 g
Temp raise (ΔT) = 10°C
Heat capacity of water (C) = 1 calorie/g-C
Formula used
Heat (ΔH) = m × C × ΔT
= 2000 × 10 = 20000 C
In Kilo calories
20000 C = 20000/1000 = 20 KCa
Hence, a food provides enough energy to raise the temperature of 2000 grams of water by 10°c contains 20 KCa.
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Answer:
One mole
Explanation:
It tells u the concentration
Answer:
0.534
Explanation:
Mole fraction can be calculated using the formula:
Mole fraction = number of moles of solute ÷ number of moles of solvent and solute (solution).
In this question, solute is dimethyl ether while the solvent is methanol.
Mole (n) = mass (M) ÷ molar mass (MM)
Mole of solute (dimethyl ether) = 148.5 ÷ 46.07
= 3.22moles.
Mole of solvent (methanol) = 90 ÷ 32.04
= 2.81moles.
Total moles of solute and solvent = 3.22 + 2.81 = 6.03moles.
Mole fraction of dimethyl ether = number of moles of dimethyl ether ÷ number of moles of solution (dimethyl ether + methanol)
Mole fraction = 3.22/6.03
= 0.534
Pressure is 5.7 atm
<u>Explanation:</u>
P1 = Standard pressure = 1 atm
P2 = ?
V1 = Volume = 10L
V2= 2.4L
T1 = 0°C + 273 K = 273 K
T2 = 100°C + 273 K = 373 K
We have to find the pressure of the gas, by using the gas formula as,
P2 can be found by rewriting the above expression as,
Plugin the above values as,
Answer:
The answer to your question is [H₃O⁺] = 0.025 [OH⁻] = 3.98 x 10⁻¹³
Explanation:
Data
[H⁺] = ?
[OH⁻] = ?
pH = 1.6
Process
Use the pH formula to calculate the [H₃O⁺], then calculate the pOH and with this value, calculate the [OH⁻].
pH formula
pH = -log[H₃O⁺]
-Substitution
1.6 = -log[H₃O⁺]
-Simplification
[H₃O⁺] = antilog (-1,6)
-Result
[H₃O⁺] = 0.025
-Calculate the pOH
pOH = 14 - pH
-Substitution
pOH = 14 - 1.6
-Result
pOH = 12.4
-Calculate the [OH⁻]
12.4 = -log[OH⁻]
-Simplification
[OH⁻] = antilog(-12.4)
-Result
[OH⁻] = 3.98 x 10⁻¹³
