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vladimir1956 [14]
4 years ago
8

An ice cube tray contains enough water at 22.0 C to make 18 ice cubes that each have a mass of 30.0g. The tray is placed in a fr

eezer that uses CF2CL2 as a refridgerant. The heat of vap of CF2CL2 is 158 J/g. What mass of [email protected] must be vaporized in the refridgeration cycle to convert all the water at 22.0 C to ice at -5.o C.
this is all i have so far. please help? thanks H capacity H20 (s) = 2.08 J/g C H capacity H20 (l) = 4.18 J/g C ethanply of fusion of ice =6.02 KJ/mol.
Chemistry
2 answers:
zloy xaker [14]4 years ago
7 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
158 J is required to vapourize 1 g of CF2Cl2
So for 235.kJ we need. 235.76 x 10^3 J/ 158 J = 1492.15g of CF2Cl2 = 1.492 kg of CF2Cl2

VMariaS [17]4 years ago
4 0

Heat capacity can be illustrated as the amount of heat absorbed or released to modify the temperature of the compound. It can be represented as, q = mcΔT

Here, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

On the basis of the given values,

Mass of one ice cube = 30 g

Mass of 10 ice cubes = 30 × 18 = 540 g

Temperature of water = 22 °C

Temperature of ice = -5 °C

Specific heat capacity of water (s) = 2.08 J/g °C

Specific heat capacity of water (l) = 4.18 J/g °C

ΔHfusion (ice) = 6.02 kJ/g

ΔHvap (CF₂Cl₂) = 158 J/g

The heat change is calculated when temperature of water changes from 22°C to 0°C,

q1 = mwater.cwater.ΔT

= 540 g × 4.18 J/g °C * (0°C -22°C)

= -49.74 kJ

Heat change when 540 g of water freezes at O°C is termed as enthalpy of fusion that is given as,

q2 = ΔHfusion × mass of water

= -6.02 kJ/mol × 540 g / 18 g/mol

= -180.6 kJ

The heat change has to be calculated when temperature of 540 g ice changes from 0°C to -5°C (Tf) as,

q3 = mice.cice.ΔT

= 540 g × 2.08 J/g°C * (-5°C - 0°C)

= -5.62 kJ

Total heat lost by the system = q1 +q2 + q3

= (49.74 + 180.6 + 5.62)

= 235.96 kJ

Now, the heat gained by CF₂Cl₂ = Heat lost by the system

ΔHvap × mass of CF₂Cl₂ vaporized = 235.96 kJ

Mass of CF₂Cl₂ vaporized = 235.96 kJ / ΔHvap

= 235.96 kJ / 158 × 10⁻³ kJ/g

= 1.5 × 10³ g

Hence, the amount of CF₂Cl₂ vaporizes in the overall conversion is 1.5 × 10³ g

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How many grams sodium bromide can be formed from 51 grams of sodium hydroxide?
raketka [301]

Explanation:

When working with moles only, you will start by applying stoichiometry to determine how the reactants will affect your amount of products in this reaction. For this question, we will assume that other reactants are in infinite qualities, so therefore, it is the amount of aluminum that we will be concerned with. You need to figure out how much aluminum is in the specified amount of aluminum bromide, and then how much aluminum hydroxide that will be able to create. Make sure all your units cancel out!

9.24 mol AlBr3 x (1 mol Al / 1 mol AlBr3) x (1 mol Al(OH)3 / 1 mol Al) = 9.24 mol AlBr3

When you're working with mole ratios that involve grams to moles conversions, the first thing you want to do is calculate the molecular weight of each component you are being asked about. Because the question was given to you as words instead of chemical formulas, you will want to figure out the chemical formulas. For example, aluminum hydroxide is Al(OH)3 and aluminum bromide is AlBr3. To calculate molecular weight, you will want to consult a periodic table, find the molecular weight for each atom, and then calculate the correct sum of each molecular weight. Make sure you keep track of the number of each atom you have, i.e. 3 oxygen and 3 hydrogen for aluminum hydroxide.

Na = 22.990 g/mol

O = 15.999 g/mol

H = 1.008 g/mol

NaOH = 22.990 g/mol + 15.999 g/mol + 1.008 g/mol = 39.997 g/mol

Al = 26.982 g/mol

O = 15.999 g/mol

H = 1.008 g/mol

Al(OH)3 = 26.982 g/mol + (3 x 15.999 g/mol) + (3 x 1.008 g/mol) = 78.003 g/mol

Now, if you begin with an amount of NaOH in grams, you will first have to convert that to moles in order to use the mole ratio.

24 g NaOH x (1 mol NaOH / 39.997 g NaOH) = 0.600 mol NaOH

Now, you will have to account for the part of the sodium hydroxide that will be present in the aluminum hydroxide. In this case, it is the hydroxide (OH) portion of the formula. There is one mole of OH in each mole of NaOH, but there are 3 moles of OH in each mole of Al(OH)3. You will start with the 0.600 mol NaOH you know you have and then use the mole ratio.

0.600 mol NaOH x (1 mol OH / 1 mol NaOH) x (1 mol Al(OH)3 / 3 mol OH) = 0.200 mol Al(OH)3

Finally, when you are converting from grams to grams, you will have to find the molecular weight of both the reactant and the product, convert reactants in grams to reactants in moles, then use the mole ratio, then convert the moles of product back to grams of product. This time, you are concerned about the mole ratio of sodium, as that is the element that is in both chemical formulas.

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2 years ago
An area where material from deep within Earth's mantle rises to the crust and melts to form magma is called a
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Answer:

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6 0
3 years ago
How to change 5 % W/V of NaCl to ppm , M ? molar mass = 58.5<br>please clear explain​
11Alexandr11 [23.1K]

Answer:

50000ppm and 0.855M.

Explanation:

ppm is an unit of chemistry defined as the ratio between mg of solute (NaCl) and Liters of solution. Molarity, M, is the ratio between moles of NaCl and liters

A 5% (w/v) NaCl contains 5g of NaCl in 100mL of solution.

To solve the ppm of this solution we need to find the mg of NaCl and the L of solution:

<em>mg NaCl:</em>

5g * (1000mg / 1g) = 5000mg

<em>L Solution:</em>

100mL * (1L / 1000mL) = 0.100L

ppm:

5000mg / 0.100L = 50000ppm

To find molarity we need to obtain the moles of NaCl in 5g using its molar mass:

5g * (1mol / 58.5g) = 0.0855moles NaCl

Molarity:

0.0855mol NaCl / 0.100L = 0.855M

7 0
3 years ago
Convert the composition of the following alloy from atom percent to weight percent (a) 44.9 at% of silver, (b) 46.3 at% of gold,
vesna_86 [32]

Answer:

Mass percentage of gold : 33.35%

Mass percentage of silver: 62.80%

Mass percentage of copper : 3.85%

Explanation:

N=n\times N_A

Where : N = Number of atoms

n = moles

N_A=6.022\times 10^{23} = Avogadro number

Let the total atoms present in the alloy be 100.

Atom percentage of silver = 44.9%

Atoms of of silver = 44.9% of 100 atoms = 44.9

Atomic weight of silver = 107.87 g/mol

Mass of 44.9 silver atoms = 107.87 g/mol ×  44.9 = 4,843.363 g/mol

Atom percentage of gold= 46.3%

Atoms of of gold = 46.3% of 100 atoms = 46.3

Atomic weight of gold = 196.97 g/mol

Mass of 44.9 gold atoms = 196.97 g/mol × 46.3 = 9,119.711 g/mol

Atom percentage of copper = 8.8 %

Atoms of of copper = 8.8% of 100 atoms = 8.8

Atomic weight of copper = 63.55 g/mol

Mass of 44.9 copper atoms = 63.55 g/mol ×  8.8 = 559.24 g/mol

Total mass of an alloy :4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol

Mass percentage of gold :

\frac{4,843.363 g/mol}{4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol}\times 100=33.35 \%

Mass percentage of silver:

\frac{9,119.711 g/mol}{4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol}\times 100=62.80\%

Mass percentage of copper :

\frac{559.24 g/mol}{4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol}\times 100=3.85\%

7 0
3 years ago
What is the current produced when a 12-Volt battery encounters a resistance of 20 Ohms?
Verdich [7]

Answer:

Explanation:

V = IR

I = V/R

I = (12V)/(20Ω)

I = 0.6 A

4 0
3 years ago
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