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2 years ago

The heat of fusion for ice is 334 joules per gram. Adding 334 joules of heat to one gram of ice at STP will

Chemistry

1 answer:

yulyashka [42]2 years ago

3 0

C) change to water at the same temperature

Explanation:

Adding 334Joules of heat to one gram of ice at STP will cause ice to change to water at the same temperature.

The heat of fusion is the amount of energy needed to melt a given mass of a solid
It is also conversely the amount of energy removed from a substance to freeze it.
The addition of this energy does not cause a decrease or increase in temperature.
Only a phase change occurs.

Learn more:

Heat of fusion brainly.com/question/4050938

#learnwithBrainly

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A generic element, Z, has two isotopes, 45Z and 47Z, and an average atomic mass of 45.36 amu. The natural abundances of the two

Dmitry_Shevchenko [17]

Answer:

46.96 amu

Explanation:

Isotopes are different kinds of same elements. The difference between two isotopes of the same element is the number of neutrons.

To get the relative atomic mass, we take into consideration the masses of the different isotopes. This is done by multiplying their abundances by their masses. They are then added together to get the relative atomic mass of the element.

Let the isotopic mass of 47Z be x

45.36 = [80/100 * 44.96] + [20/100 * x]

45.36 = 35.968 + 0.2x

0.2x = 45.36 - 35.968

0.2x = 9.392

x =9.392/0.2 = 46.96 amu

5 0

2 years ago

A scientist makes an acid solution by adding drops of acid to 1.2 l of water. the final volume of the acid solution is 1.202 l.

Nataliya [291]

Answer:- 40 drops and the percentage of acid in acid solution is 0.166%.

Solution:- Acid is added drops wise to 1.2 L of water to make acid solution. The final volume of the acid solution is 1.202 L. From here we could calculate the volume of the acid added to water by subtracting water volume from final acid solution volume as:

volume of acid added = 1.202 L - 1.2 L = 0.002 L

It's given that the volume of each drop is 0.05 mL. let's convert the volume of acid added from L to mL.

$0.002L(\frac{1000mL}{1L})$

= 2mL

Now let's calculate the number of drops of acid added to water as:

$2mL(\frac{1drop}{0.05mL})$

= 40 drops

0.002 L of acid are present in 1.202 L of solution. It asks to calculate about what percent of acid solution is acid means how many L of acid are present in 100 L of acid solution. It's done as:

percentage of acid in acid solution = $(\frac{0.002}{1.202})100$

= 0.166%

So, 40 drops of acid are required and there is 0.166% of acid in acid solution.

5 0

2 years ago

1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl

kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0

2 years ago

A 0.271g sample of an unknown vapor occupies 294ml at 140C and 874mmHg. The emperical formula of the compound is CH2. How many m

Phoenix [80]

Using PV = nRT, we can calculate the moles of the sample.
874 mmHg = 116,524 Pa
n = PV/RT
n = 116,524 x 294 x 10⁻⁶ / 8.314 x (140 + 273)
n = 9.98 x 10⁻³ mol

moles = mass / Mr
Mr = 0.271/9.98 x 10⁻³
Mr = 27.2
Mass of empirical formula = 14
Repeat units = 27.2 / 14 ≈ 2

Formula of substance:
C₂H₄

Combustion equation:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

1 mole produces 2 moles of CO₂, so 3 moles will produce 6 moles CO₂

7 0

2 years ago

Ricardo finds an online site about the gas laws. The site shows the equation below for Charles’s law.

Sati [7]

Answer:The symbol for T2 should be smaller than for T1 because if volume increases, then temperature should decrease.

Explanation:

4 0

2 years ago

Read 2 more answers