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2 years ago

A megamole of titanium contains __________ moles of titanium.

2 answers:

8 0

Answer is: 1 megamole is equal to 1000000 (one million) mole or 10⁶ moles.
The SI base unit for amount of substance is the mole. The SI prefix "mega" represents a factor of 10⁶, or in exponential notation, 1E6.
The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 12 grams of carbon C-12.

vlada-n [284]2 years ago

7 0

Answer : A megamole of titanium contains 1000000 moles of titanium.

Explanation :

As we are given:

Number of moles of titanium is 1 megamole.

The conversion used from megamole to mole is:

1 megamole = 1000000 mole

or,

1 megamole = $10^6$ mole

Hence, 1 megamole of titanium contains 1000000 moles of titanium.

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A 40.0 mL sample of 0.25 M KOH is added to 60.0 mL of 0.15 M Ba(OH)2. What is the molar concentration of OH-(aq) in the resultin

solmaris [256]

Answer:

C) 0.28 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Potassium hydroxide will furnish hydroxide ions as:

$KOH\rightarrow K^{+}+OH^-$

Given :

For Potassium hydroxide :

Molarity = 0.25 M

Volume = 40.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 40.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Potassium hydroxide is same as the moles of Potassium hydroxide as shown below:

$Moles =0.25 \times {40.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Potassium hydroxide = 0.01 moles

Barium hydroxide will furnish hydroxide ions as:

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^-$

Given :

For Barium hydroxide :

Molarity = 0.15 M

Volume = 60.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 60.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Barium hydroxide is twice the moles of Barium hydroxide as shown below:

$Moles =2\times 0.15 \times {60.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Barium hydroxide = 0.018 moles

Total moles = 0.01 moles + 0.018 moles = 0.028 moles

Total volume = 40.0×10⁻³ L + 60.0×10⁻³ L = 100×10⁻³ L

Concentration of hydroxide ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{OH^-}=\frac{0.028 }{100\times 10^{-3}}$

The final concentration of hydroxide ion = 0.28 M

5 0

2 years ago

Read 2 more answers

What volume of 0.210 M sulfuric acid is required to completely react with 2.14 g aluminium hydroxide?

Galina-37 [17]

3 H2SO4 + 2 Al(OH)3 → Al2(SO4)3 + 6 H2O

(2.14 g Al(OH)3) / (78.0036 g Al(OH)3/mol) x (3 mol H2SO4 / 2 mol Al(OH)3) / (0.210 mol/L H2SO4) =
0.19596 L = 196 mL H2SO4

3 0

2 years ago

A compound that is composed of carbon, hydrogen, and oxygen contains 70.6% C, 5.9% H, and 23.5% O by mass. The molecular weight

zhannawk [14.2K]

Answer: The molecular formula will be $C_8H_8O_2$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 70.6 g

Mass of H = 5.9 g

Mass of O = 23.5 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{70.6g}{12g/mole}=5.9moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.9g}{1g/mole}=5.9moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.5g}{16g/mole}=1.5moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{5.9}{1.5}=4$

For H = $\frac{5.9}{1.5}=4$

For O = $\frac{1.5}{1.5}=1$

The ratio of C : H: O= 4: 4:1

Hence the empirical formula is $C_4H_4O$

The empirical weight of $C_4H_4O$ = 4(12)+4(1)+1(16)= 68g.

The molecular weight = 136 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{136}{68}=2$

The molecular formula will be= $2\times C_4H_4O=C_8H_8O_2$

4 0

2 years ago

1. A student sees tiny bubbles clinging to the inside of an unopened plastic bottle full of carbonated soft drinks. The student

Lina20 [59]

Answer:

1) The bubbles will grow, and more may appear.

2)Can A will make a louder and stronger fizz than can B.

Explanation:

When you squeeze the sides of the bottle you increase the pressure pushing on the bubble, making it compress into a smaller space. This decrease in volume causes the bubble to increase in density. When the bubble increases in density, the bubble will grow and more bubbles will appear. Therefore, Changing the pressure (by squeezing the bottle) changes the volume of the bubbles. The number of bubbles doesn't change, just their size increases.

Carbonated drinks tend to lose their fizz at higher temperatures because the loss of carbon dioxide in liquids is increased as temperature is raised. This can be explained by the fact that when carbonated liquids are exposed to high temperatures, the solubility of gases in them is decreased. Hence the solubility of CO2 gas in can A at 32°C is less than the solubility of CO2 in can B at 8°C. Thus can A will tend to make a louder fizz more than can B.

3 0

2 years ago

How many grams of Cr can be produced by the reaction of 44.1 g of Cr2O3 with 35.0 g of Al according to the following chemical eq

allsm [11]

Answer:

30.17 grams of Cr can be produced by the reaction of 44.1 g of Cr₂O₃ with 35.0 g of Al.

19.33 grams of the excess react will remain once the reaction goes to completion.

Explanation:

You know:

2 Al + Cr₂O₃ → Al₂O₃ + 2 Cr

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of the compounds participating in the reaction react and are produced:

Al: 2 moles
Cr₂O₃: 1 mole
Al₂O₃: 1 mole
Cr: 2 moles

Being:

Al: 27 g/mole
Cr: 52 g/mole
O: 16 g/mole

the molar mass of the compounds participating in the reaction is:

Al: 27 g/mole
Cr₂O₃: 2*52 g/mole + 3 *16 g/mole= 152 g/mole
Al₂O₃: 2*27 g/mole + 3 *16 g/mole= 102 g/mole
Cr: 52 g/mole

Then, by stoichiometry of the reaction, the amounts of reagents and products that participate in the reaction are:

Al: 2 moles* 27 g/mole= 54 g
Cr₂O₃: 1 mole* 152 g/mole= 152 g
Al₂O₃: 1 mole* 102 g/mole= 102 g
Cr: 2moles*52 g/mole= 104 g

First, you must determine the limiting reagent. The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

Then, for this you must apply the following rule of three: if 54 grams of Al react with 152 grams of Cr₂O₃ by stoichiometry of the reaction, 35 grams of Al with how much mass of Cr₂O₃ will react?

$mass of Cr_{2} O_{3} =\frac{35 grams of Al*152 gramsof Cr_{2} O_{3}}{54 grams of Al}$

mass of Cr₂O₃= 98.52 grams

But 98.52 grams of Cr₂O₃ are not available, 44.1 grams are available. Since you have less moles than you need to react with 35 grams of Al, the compound Cr₂O₃ will be the limiting reagent.

To determine the amount of Cr that can be produced, you use the amount of limiting reagent available and apply the following rule of three: if by stoichiometry 152 grams of Cr₂O₃ produce 104 grams of Cr, 44.1 grams of Cr₂O₃ how much mass of Cr does it produce?

$mass of Cr =\frac{104 grams of Cr*44.1 grams of Cr_{2} O_{3}}{152 grams of Cr_{2} O_{3}}$

mass of Cr= 30.17 grams

30.17 grams of Cr can be produced by the reaction of 44.1 g of Cr₂O₃ with 35.0 g of Al.

As Al is the excess reagent, you must first calculate the amount of mass that reacts with 44.1 grams of Cr₂O₃ using the following rule of three: if 152 grams of Cr₂O₃ react with 54 grams of Al, 44.1 grams of Cr₂O₃ with how much mass of Al reaction to?

$mass of Al =\frac{54 grams of Al*44.1 gramsnof Cr_{2} O_{3}}{152 gramsnof Cr_{2} O_{3}l}$

mass of Al=15.67 grams

With 35 grams being the amount of Al available, the amount of Al that will remain in the reaction after all the limiting reagent reacts and the reaction is complete is calculated by:

mass of excess= 35 grams - 15.67 grams= 19.33 grams

19.33 grams of the excess react will remain once the reaction goes to completion.

6 0

2 years ago