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igomit [66]
3 years ago
13

The conversion of 1 mole of water at 0°c to 1 mole of ice at 0°c releases 6.01 kj. what must be true about the heat absorbed dur

ing the melting of 1 mole of ice at 0°c to 1 mole of water at 0°c? it must be less than 6.01 kj. it must be greater than 6.01 kj. it must be 6.01 kj.
Chemistry
2 answers:
Stels [109]3 years ago
7 0

Answer: it must be 6.01 kJ.

Explanation:

Latent heat of freezing is the amount of heat released to convert 1 mole of liquid to 1 mole of solid at atmospheric pressure.

H_2O(l)\rightarrow H_2O(s)

Given: latent heat of freezing is 6.01 kJ.

Latent heat of fusion is the amount of heat required to convert 1 mole of solid to 1 mole of liquid at atmospheric pressure. If latent heat of freezing is 6.01 kJ, latent heat of fusion will be same but the heat will be absorbed.

H_2O(s)\rightarrow H_2O(l)

love history [14]3 years ago
6 0
It must be 6.01 kj
................



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If 1.02 g of nickel reacted with 750. mL of 0.112 M hydrobromic acid, how much of each will be present at the end of the reactio
kati45 [8]

Answer:

35.1% is percent yield

Explanation:

<em>Full question: Assume no volume change.  If you formed 0.0910 atm of gas, what is the percent yield?</em>

<em />

The reaction that is occurring is:

Ni + 3HBr → NiBr₃ + 3/2H₂(g)

First, we will determine moles of Ni and HBr to determine limiting reactant and theoretical yield

Using ideal gas law, we can determine the moles of hydrogen formed. Thus, we can find percent yield:

<em>Moles Ni (Molar mass: 58.69g/mol):</em>

1.02g * (1mol / 58.69g) = 0.01738moles Ni

<em>Moles HBr:</em>

0.750L * (0.112mol/L) = 0.084 moles of HBr.

For a complete reaction of the 0.084 moles of HBr you need:

0.084mol HBr * (1 mole Ni / 3 moles HBr) = 0.028 moles of Ni.

As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

0.01738moles Ni * (3/2 H₂ / 1 mol Ni) = 0.02607 moles H₂

Now, moles of H₂ produced are:

PV = nRT

PV/RT = n

<em>Where P is pressure (0.0910atm)</em>

<em>V is volume (2.50L)</em>

<em>R is gas constant (0.082atmL/molK)</em>

<em>T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)</em>

<em>And n are moles</em>

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

<em />

And percent yield (Produced moles / Theoretical moles * 100) is:

0.00915 moles / 0.02607moles =

<h3>35.1% is percent yield</h3>
8 0
2 years ago
What is the molarity of 5.60 mol of sodium carbonate in 1500 ml of solution?
FrozenT [24]

Answer:

3.74 M

Explanation:

We know that molarity is moles divided by liters. The first thing to do here is convert your 1500 mL of solution to L. There's 1,000 mL in 1 L, so you need to divide 1500 by 1000:

1500 ÷ 1000 = 1.50

Now you can plug your values into the equation for molarity:

5.60 mol ÷ 1.50 L = 3.74 M

7 0
3 years ago
The stomach and intestines are organs of the body system which _______.
yulyashka [42]

Answer:

A. Digests food

Explanation:

Look about diggestive process in Google

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3 0
3 years ago
Read 2 more answers
Compare the amount of heat required to vaporize a 200.-gram sample of H2O(ℓ) at its boiling point to the amount of heat required
stira [4]
The enthalpy of vaporization of H2O is higher than the enthalpy of fusion of H2O, therefore vaporizing the same mass of H2O would require more heat/energy than melting the same mass of H2O.
6 0
3 years ago
How many moles of aspartame are present in 1.00 mg of aspartame?
Diano4ka-milaya [45]

Answer:- There are 3.40*10^-^6 moles.

Solution:- It is a unit conversion problem where we are asked to convert mg of aspartame to moles. Aspartame is C_1_4H_1_8N_2O_5 and it's molar mass is 294.31 grams per mole.

mg are converted to grams and then the grams are converted to moles as:

1.00mg Aspartame(\frac{1g}{1000mg})(\frac{1mole}{294.31g})

= 3.40*10^-^6 moles of aspartame

So, there would be 3.40*10^-^6 moles of aspartame in 1.00 mg of it.

3 0
3 years ago
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