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Hatshy [7]
3 years ago
7

K12 3.10 Unit Assessment: Solutions, Part 1 does anyone have the answers for this quiz

Chemistry
2 answers:
Pani-rosa [81]3 years ago
5 0
Eat eat eat eat eat eat eat eat eat
Stells [14]3 years ago
3 0

Answer:

no i was looking for them can you help

Explanation:

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Which element is likely to be the best conductor of electricity?
Eddi Din [679]

Answer:

copper

Explanation:

4 0
3 years ago
What is the difference between reactants and products?
vitfil [10]

Answer:

The reactants are the substances that start the chemical reaction. The products are the substances that are produced in the chemical reaction.

Explanation:

5 0
3 years ago
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How many moles of agcl are contained in 244 ml of 0.135 magcl solution? the density of the solution is 1.22 g/ml?
spin [16.1K]
Since we are only asked for the number of moles, we don't need the information of density. The concentration is expressed in terms of 0.135 M AgCl or 0.135 moles of AgCl per liter solution. The solution is as follows:

Moles AgCl = Molarity * Volume
Moles AgCl = 0.135 mol/L * 244 mL * 1 L/1000 mL
<em>Moles AgCl = 0.03294 mol </em>
3 0
3 years ago
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The enthalpy of fusion of solid n-butane is 4.66 kJ/mol. Calculate the energy required to melt 58.3 g of solid n-butane.
adelina 88 [10]

Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

Explanation :

First we have to calculate the moles of n-butane.

\text{Moles of n-butane}=\frac{\text{Mass of n-butane}}{\text{Molar mass of n-butane}}

Given:

Molar mass of n-butane = 58.12 g/mole

Mass of n-butane = 58.3 g

Now put all the given values in the above expression, we get:

\text{Moles of n-butane}=\frac{58.3g}{58.12g/mol}=1.00mol

Now we have to calculate the energy required.

Q=\frac{\Delta H}{n}

where,

Q = energy required

\Delta H = enthalpy of fusion of solid n-butane = 4.66 kJ/mol

n = moles = 1.00 mol

Now put all the given values in the above expression, we get:

Q=\frac{4.66kJ/mol}{1.00mol}=4.66kJ

Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

7 0
3 years ago
How much energy is used to melt 44.33 g of solid oxygen?
Nutka1998 [239]

Answer:

Q1 = C * m * dT

Q2 = Qm * m

Qtotal = Q1 + Q2

Q1 - is amount of energy you need to apply to heat oxygen from the current temperature till you reach the melting temperature. Only if the oxygen is below to melting temperature.

C - is calorific capacity of oxygen -- better look at tables, it is a constant value

m - is the amount of oxygen, we will use moles because the other data shows moles, but could be grams, kg, etc.

dT - is the diference of temperatures between the current and the melting one. The melting temperature is constant and you can find it on tables, then (Tm - To)

Q2 is the amount of energy you have to add to melt oxygen once the oxygen has reached the melting temperature (Tm)

Qm is a constant value you could find on tables, depends on the mass of oxygen and is due to internal processes as changes in atomic distributions

If the oxygen is initially at melting temperature (melting point) you only need to know Q2, as dT = 0

I will do an example for you, but in future you should provide data of constants, it takes very long to find them in books or internet.

Data from tables

Tm =  54.36 K

C = 29.378 J/mol K this is at 25 C (or 298 K), is not really correct, you should look at its value at less than 54.36 K, but you can use it here.

Qm = 0.444 kJ/mol

Problem -- you have 44.33g of Oxygen -- Molecular weight of O2 is 32 g/mol

So you have 44.33/32 = 1.385 moles of oxygen

a) if oxygen is already at melting temperature: you only have to melt it

Qtotal = Q1 + Q2 = [0 (dT = 0) + Qm * m] = 0.444 * 1.385 = 0.615 kJ = 615 J

b) supposing an initial temperture of 50 K: now you have to heat oxygen till melting temperature and then melt it.

Q1 = C * m * dT = 29.378 * 1.385 * (54.36 - 50) = 177.442 J

Q2 = Qm * m = 615 J

Qtotal = 177.442 + 615 = 792.44 J

Explanation:

4 0
3 years ago
Read 2 more answers
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