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2 months ago

**Answer:**

false

**Explanation:**

Question

Overall mineral crystal size depends on how fast a solution _____.

Solution 1

In overall, the size of the mineral crystal depends on how fast a solution cools down. More time that is spent during cooling creates larger crystals because of the free movement of the constituent of mineral isotopesÂ in a somewhat liquid solution.Â

Solution 2

Answer:cools

Explanation:

Question

Natural gas is primarily composed of ________. natural gas is primarily composed of ________. methane sulfur dioxide nitrogen oxygen nitrite

Solution 1

NaturalÂ gasÂ is primarily composed ofÂ Â **methane (CH4)**

Â NaturalÂ gasÂ is a naturallyÂ occurringÂ hydrocarbonÂ mixtureÂ whichÂ isÂ primarilyÂ composed of Methane(CH4),Â but itÂ alsoÂ contains ethane,propane andÂ heavier hydrocarbon. In additionÂ itÂ contain small amountÂ of nitrogen, carbon dioxide,hydrogen sulfide and traces amount of water.

Â N

Question

It is necessary to have a 40% antifreeze solution in the radiator of a certain car. the radiator now has 70 liters of 20% solution. how many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?

Solution 1

The answer is : 1**7.5 liters drained and replaced by 17.5 liters of 100% solution.Â **

x = amount drained and replaced

(70-x) = remaining amount of 20% solution

.20(70-x) + 1.00(x) = .40(70)

14 - .2x + 1x = 28

1x - .2x = 28 - 14

.8x = 14

x = 14/.8

x= 17,5 ( 17.5 liters drained and replaced by 17.5 liters of 100% solution)

(70-x) = remaining amount of 20% solution

.20(70-x) + 1.00(x) = .40(70)

14 - .2x + 1x = 28

1x - .2x = 28 - 14

.8x = 14

x = 14/.8

x= 17,5 ( 17.5 liters drained and replaced by 17.5 liters of 100% solution)

Solution 2

**Answer:**

17.5 litres removed and 17.5 litres of pure antifreeze added

**Explanation:**

Let k equal the amount of the solution to be removed

amount of antifreeze to be added

0.2(70 - x) + x = 0.4(70)

14 - 0.2x +x = 28

0.8x = 28 -14

x = 14/0.8 = 17.5

x = 17.5 litres removed and 17.5 litres of pure antifreeze added

Question

How many kilojoules are required to convert 115.0 g of ice at 0.0 âˆ˜c to liquid water at 32 âˆ˜c? the heat of fusion of water is 334 j/g, and the heat capacity of water is 4.184 j/g âˆ˜c?

Solution 1

The answer is **53.8 kJ**.

Solution:There are two major steps in converting ice to liquid water. It begins with a phase change when ice melts at 0.0Â°C, and then a temperature change when the liquid water rises inÂ temperature from zero to 32Â°C.

The amount of heat involved with the phase change melting is given by

Â Â Â q = (mass of water) (Î”Hfus)

Â Â Â Â = (115.0 g)(334 J/g)Â

Â Â Â Â = 38410 J = 38.41 kJ

The amount of heat involved with temperature change isÂ

Â Â Â q = mcÎ”T

Â Â Â Â = (115.0g)(4.184J/gÂ°C)(32Â°C - 0.0Â°C)

Â Â Â Â = 15397.12 J = 15.39712 kJ

Summing up the two values gives the total heat required to convert ice to liquid water:

Â Â Â q = 38.41 kJ + 15.39712 kJ= 53.8 kJ

Solution:There are two major steps in converting ice to liquid water. It begins with a phase change when ice melts at 0.0Â°C, and then a temperature change when the liquid water rises inÂ temperature from zero to 32Â°C.

The amount of heat involved with the phase change melting is given by

Â Â Â q = (mass of water) (Î”Hfus)

Â Â Â Â = (115.0 g)(334 J/g)Â

Â Â Â Â = 38410 J = 38.41 kJ

The amount of heat involved with temperature change isÂ

Â Â Â q = mcÎ”T

Â Â Â Â = (115.0g)(4.184J/gÂ°C)(32Â°C - 0.0Â°C)

Â Â Â Â = 15397.12 J = 15.39712 kJ

Summing up the two values gives the total heat required to convert ice to liquid water:

Â Â Â q = 38.41 kJ + 15.39712 kJ= 53.8 kJ

Question

0.50 g Al equals how many moles

Solution 1

**Answer: The number of moles of 0.50 g Aluminium is 0.0185 moles.**

**Explanation:**

**To calculate the number of moles, we use the formula:**

**In the given question:**

Given mass of Aluminium = 0.50 g

Molar mass of Aluminium = 27 g/mol

**Putting values in above equation, we get:**

**Hence, the number of moles of 0.50 g of Aluminium is 0.0185 moles.**

Solution 2

Molar mass of aluminum ( Al ) = 26.98 g/mol

n = m / M

n = 0.50 Â / 26.98Â

**n = 0.01853 moles**

hope this helps!

n = m / M

n = 0.50 Â / 26.98Â

hope this helps!

Question

2.50 g CuCl2 equals how many moles

Solution 1

The molar mass of CuCl2 is 134.45 g/mol; therefore, you divide 2.5 g of CuCl2 by 134.45 g of CuCl2 leaving you with 0.019 moles

Solution 2

**Explanation:**

Number of moles is defined as the mass of substance given in grams divided by the molar mass of substance.

It is given that mass of is 2.50 g and molar mass of is 134.45 g/mol.

Therefore, calculate number of moles as follows.

Â Â Â Number of moles of

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.0186 mol

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.019 mol (approx)

Thus, we can conclude that **2.50 g CuCl2 equals 0.019 moles (approx). Â Â Â Â Â Â ** Â Â Â

Question

A chemical supply company sells a concentrated solution of aqueous h2so4 (molar mass 98 g molâˆ’1 ) that is 50. percent h2so4 by mass. at 25Â°c, the density of the solution is 1.4 g mlâˆ’1 . what is the molarity of the h2so4 solution at 25Â°c?

Solution 1

Answer is:Â the molarity of the sulfuric acid is 7.14 M.

Ï‰(Hâ‚‚SOâ‚„) = 50%Â Ã· 100% = 0.5.

d(Hâ‚‚SOâ‚„) = 1.4 g/mL.

V(Hâ‚‚SOâ‚„) = 100 mL Ã· 1000 mL/L = 0.1 L..

mr(Hâ‚‚SOâ‚„) = d(Hâ‚‚SOâ‚„)Â Â· V(Hâ‚‚SOâ‚„).

mr(Hâ‚‚SOâ‚„) = 1.4 g/mLÂ Â· 100 mL.

mr(Hâ‚‚SOâ‚„) = 140 g.

m(Hâ‚‚SOâ‚„) =Â Ï‰(Hâ‚‚SOâ‚„)Â Â· mr(Hâ‚‚SOâ‚„).

m(Hâ‚‚SOâ‚„) = 0.5Â Â· 140 g.

m(Hâ‚‚SOâ‚„) = 70 g.

n(Hâ‚‚SOâ‚„) = m(Hâ‚‚SOâ‚„)Â Ã· M(Hâ‚‚SOâ‚„).

n(Hâ‚‚SOâ‚„) = 70 gÂ Ã· 98 g/mol.

n(Hâ‚‚SOâ‚„) = 0.714 mol.

c(Hâ‚‚SOâ‚„) = n(Hâ‚‚SOâ‚„)Â Ã· V(Hâ‚‚SOâ‚„).

c(Hâ‚‚SOâ‚„) = 0.714 molÂ Ã· 0.1 L.

C(Hâ‚‚SOâ‚„) = 7.14 M.

Ï‰(Hâ‚‚SOâ‚„) = 50%Â Ã· 100% = 0.5.

d(Hâ‚‚SOâ‚„) = 1.4 g/mL.

V(Hâ‚‚SOâ‚„) = 100 mL Ã· 1000 mL/L = 0.1 L..

mr(Hâ‚‚SOâ‚„) = d(Hâ‚‚SOâ‚„)Â Â· V(Hâ‚‚SOâ‚„).

mr(Hâ‚‚SOâ‚„) = 1.4 g/mLÂ Â· 100 mL.

mr(Hâ‚‚SOâ‚„) = 140 g.

m(Hâ‚‚SOâ‚„) =Â Ï‰(Hâ‚‚SOâ‚„)Â Â· mr(Hâ‚‚SOâ‚„).

m(Hâ‚‚SOâ‚„) = 0.5Â Â· 140 g.

m(Hâ‚‚SOâ‚„) = 70 g.

n(Hâ‚‚SOâ‚„) = m(Hâ‚‚SOâ‚„)Â Ã· M(Hâ‚‚SOâ‚„).

n(Hâ‚‚SOâ‚„) = 70 gÂ Ã· 98 g/mol.

n(Hâ‚‚SOâ‚„) = 0.714 mol.

c(Hâ‚‚SOâ‚„) = n(Hâ‚‚SOâ‚„)Â Ã· V(Hâ‚‚SOâ‚„).

c(Hâ‚‚SOâ‚„) = 0.714 molÂ Ã· 0.1 L.

C(Hâ‚‚SOâ‚„) = 7.14 M.

Question

Calculate the number of moles of naoh present in 11.2 ml of 2.50 m naoh solution

Solution 1

Let's review what is given in this problem.

The volume of the solution is 11.2 milliliters, which is 0.0112 liters (multiply 11.2 mL with 1 L / 1000 mL to get 0.0112).

The molar concentration is 2.5 M. This means that there are 2.5 moles of NaOH per liter in the solution.

Multiply the volume and the molar concentration to get the moles of NaOH in the solution.

The "Liter" will cancel out, leaving only moles.

After using a calculator (or computing by hand), you should get the following value:

Since the given values were given with 3 significant figures, let's change this answer so there are 3 significant figures.

Thus, your final answer is

The volume of the solution is 11.2 milliliters, which is 0.0112 liters (multiply 11.2 mL with 1 L / 1000 mL to get 0.0112).

The molar concentration is 2.5 M. This means that there are 2.5 moles of NaOH per liter in the solution.

Multiply the volume and the molar concentration to get the moles of NaOH in the solution.

The "Liter" will cancel out, leaving only moles.

After using a calculator (or computing by hand), you should get the following value:

Since the given values were given with 3 significant figures, let's change this answer so there are 3 significant figures.

Thus, your final answer is

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