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3 years ago

What equals the molar mass of a element

1 answer:

galina1969 [7]3 years ago

7 0

Molar mass is the given substance divided by the amount of that substance, measured in g/mol.

Example: titanium atomic mass is 47.88 amu or 47.88 g/mol. In 47.88 grams of titanium there is 1 mole, or 6.022 x 10^23 titanium atoms.

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How many U.S. gallons are there in a cubic mile? The total proven oil reserves of the U.S. are roughly 30 x 10°bbl. How many cub

Ira Lisetskai [31]

Answer:

1 cubic mile = 1.101 * 10^12 US gallons

1 US bbl oil = 42 US gallons = 3.8143*10^ -11 cubic miles

Explanation:

The number of the exponent of the oil reserve is not very well shown in the question so, I provide you the conversion of bbl oil into cubic mile, the only thing you have to do is multiply the number of bbls of the reserve for the conversion in cubic miles and you'll have the answer.

4 0

3 years ago

What is the mass of six of these marbles? What is the volume? What is the<br> density?

chubhunter [2.5K]

Answer:

All right. So let's calculate the density of a glass marble. Remember that the formula for density is mass over volume. So if I know that the masses 18.5 g. And I know that the um volume is 6.45 cubic centimeters. I can go ahead and answer this to three significant figures. So it's going to be 2.87 grams per cubic centimeter. Okay, that's our density. Now, density is an intensive process. Okay. We're an intensive property. I really should say. It doesn't depend on how much you have. Mhm. If I have one marble, its density is going to be 2.87 g per cubic centimeter. If I have two marbles, the density will be the same because I'll double the mass and I'll also double the volume. So when I divide them I'll get the same number. Okay, that's what makes it an intensive property. No matter how many marbles I have, they'll have the same density. Mass though is not an intensive property. So if I have six marbles and I want to know what the massive six marbles is. Well, I know the mass of each marble is 18.5 g. So the mass of six marbles Is going to be 100 11 g. Because mass is an extensive property. It depends on how much you have. If I change the number of marbles, I'm going to change the mass. That's an extensive property. All right. So we've calculated the density. We've calculated the mass and then what happens to the density of one marble compared to six marbles as we mentioned before. Since densities and intensive property, the densities will be the same, no matter how may.

Explanation:

5 0

3 years ago

Need to complete the chart

GalinKa [24]

Answer:

Row 1

$[H^+]=1.8\times 10^{-6}M$

$pH=-\log[H^+]=-\log[1.8\times 10^{-6}]=5.7$

pOh=14-pH=14-5.7=8.3

$pOH=-\log[OH^-]$

$[OH^-]=0.5\times 10^{-8}M$

Hence, acidic

Row 2

$[OH^-]=3.6\times 10^{-10}M$

$pOH=-\log[OH^-]=-\log[3.6\times 10^{-10}]=9.4$

pH=14-pOH=14 - 9.4 = 4.6

$pH=-\log[H^+]$

$[H^+]=2.6\times 10^{-5}M$

Hence, acidic

Row 3

pH = 8.15

$[H^+]=0.7\times 10^{-8}M$

pOH=14-pH=14 - 8.15 = 5.8

$pOH=-\log[OH^-]$

$[OH^-]=1.5\times 10^{-6}M$

Hence, basic

Row 4

pOH = 5.70

$[OH^-]=1.8\times 10^{-6}M$

pH=14-pOH=14 - 5.70 = 8.3

$pH=-\log[H^+]$

$[H^+]=0.5\times 10^{-8}M$

Hence, basic

3 0

3 years ago

The combustion of 14 grams of co according to the reaction co(g) ½o2(g) co2(g) 67.6 kcal gives off how much heat?

Irina18 [472]

The balanced reaction that describes the reaction between carbon monoxide and oxygen to produce carbon dioxide is expressed co(g)+ ½o2(g) <span>→ </span>co2(g). The heat of combustion is 67.6 kcal per mole CO. We are given with 14 grams CO. Convert this to mole equal to 0.5 mole. Hence the total heat is 33.8 kcal.

3 0

3 years ago

The mass of sodium chloride in (g) is 14.19 The volume of ammonia solution in (mL) is 36.15 Calculate the following: What is the

Svetach [21]

This is an incomplete question, here is a complete question.

A weighed amount of sodium chloride is completely dissolved in a measured volume of 4.00 M ammonia solution at ice temperature, and carbon dioxide is bubbled in. Assume that sodium bicarbonate is formed util the limiting reagent is entirely used up. the solubility of sodium bicarbonate in water at ice temperature is 0.75 mol per liter. Also assume that all the sodium bicarbonate precipitated is collected and converted quantitatively to sodium carbonate.

Data to be used for calculating the results

-The mass of sodium chloride in (g) is 14.19

-The volume of ammonia solution in (mL) is 36.15

Calculate the following: What is the theoretical yield of sodium bicarbonate in grams?

Answer : The theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

Explanation :

First we have to calculate the moles of NaCl and $NH_3$ .

$\text{ Moles of }NaCl=\frac{\text{ Mass of }NaCl}{\text{ Molar mass of }NaCl}=\frac{14.19g}{58.5g/mole}=0.243moles$

$\text{ Moles of }NH_3=\text{ Concentration of }NH_3\times \text{ Volume of solution}=4.00M\times 0.3615L=1.446moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$NH_3+NaCl+CO_2+H_2O\rightarrow NaHCO_3+NH_4Cl$

From the balanced reaction we conclude that

As, 1 mole of $NaCl$ react with 1 mole of $NH_3$

So, 0.243 mole of $NaCl$ react with 0.243 mole of $NH_3$

From this we conclude that, $NH_3$ is an excess reagent because the given moles are greater than the required moles and $NaCl$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaHCO_3$

From the reaction, we conclude that

As, 1 mole of $NaCl$ react to give 1 mole of $NaHCO_3$

So, 0.243 moles of $NaCl$ react to give 0.243 moles of $NaHCO_3$

Now we have to calculate the mass of $NaHCO_3$

$\text{ Mass of }NaHCO_3=\text{ Moles of }NaHCO_3\times \text{ Molar mass of }NaHCO_3$

Molar mass of sodium bicarbonate = 84 g/mol

$\text{ Mass of }NaHCO_3=(0.243moles)\times (84g/mole)=20.4g$

Thus, the theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

4 0

3 years ago