Answer: 2800 calories
Explanation:
Latent heat of fusion is the amount of heat required to convert 1 mole of solid to liquid at atmospheric pressure.
Amount of heat required to fuse 1 gram of water = 80 cal
Mass of ice given = 35 gram
Heat required to fuse 1 g of ice at
= 80 cal
Thus Heat required to fuse 35 g of ice =
Thus 2800 calories of energy is required to melt 35 g ice cube
Answer:
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Explanation:
<u>1. Chemical balanced equation (given)</u>

<u>2. Mole ratio</u>

This is, 1 mol of NaOH will reacts with 1 mol of KHP.
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<u>3. Find the number of moles in 72.14 mL of the base</u>



<u>4. Find the number of grams of KHP that reacted</u>
The number of moles of KHP that reacted is equal to the number of moles of NaOH, 0.007055 mol
Convert moles to grams:
- mass = number moles × molar mass = 0.007055mol × 204.23g/mol
You have to round to 3 significant figures: 1.44 g (because the molarity is given with 3 significant figures).
<u>5. Find the percentage of KHP in the sample</u>
The percentage is how much of the substance is in 100 parts of the sample.
The formula is:
- % = (mass of substance / mass of sample) × 100
- % = (1.4408g/ 1.864g) × 100 = 77.3%
Answer:
A Bronsted-Lowry acid like and Arrhenius acid is a compound that breaks down to give an H+ in solution. The only difference is that the solution does not have to be water. ... An Arrhenius base is a molecule that when dissolved in water will break down to yield an OH- or hydroxide in solution.
Explanation:
Answer:
during the process of carbon cycle
Answer:
The balloon will occupy a volume of 6.48 L
Explanation:
<u>Step 1: </u>Data given
internal pressure = 1.00 atm
volume = 4.50 L
Temperature = 20.0 °C
<u>Step 2:</u> Calculate new volume via the ideal gas law
P*V = n*R*T
(P1*V1)/ T1 = (P2*V2)/T2
⇒ with P1 = 1.00 atm
⇒ with V1 = 4.50 L
⇒ with T1 = 20.0 °C = 293 Kelvin
⇒ with P2 = 0.600 atm
⇒ with V2 = TO BE DETERMINED
⇒ with T2 = -20°C = 253 Kelvin
(1.00atm * 4.50 L)/293 Kelvin = (0.600 atm*V2) / 253 Kelvin
0.01536 = 0.00237 V2
V2 = 6.48 L
The balloon will occupy a volume of 6.48 L