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vlabodo [156]
3 years ago
7

What would the total mass of a 4.9 mole sample of Barium (Ba) be if its Molar Mass is 137g?

Chemistry
1 answer:
Aleksandr-060686 [28]3 years ago
8 0

Answer:

m = 671 grams

Explanation:

Given that,

No of moles, n = 4.9

Molar mass of Barium, M = 137 g

Mass divided by molar mass is equal to no of moles. It can be given by the formula as follows :

n=\dfrac{m}{M}\\\\m=n\times M\\\\m=4.9\times 137\\\\m=671.3\ g

or

m = 671 grams

So, the total mass of the sample of Barium is 671 grams.

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The specific heat of aluminum is 0.897j/(g•°C). If a 22.6 g sample of aluminum is heated from 25°C to 250°C, then how much heat
Oxana [17]

Q=mcat

=(22.6)(.897)(250-25)

4,561.245 or 4.5 * 10^-3

3 0
3 years ago
How many degrees of unsaturation are lost or gained in the base-induced transformation from mito-ph to its photoactive ring-open
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Unsaturation (IHD) 2 hydrogen Needed

IHD = [(2n+2) -H]/2
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Unsaturation:
Double bonds = 1
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Triple Bonds = 2

The degrees of unsaturation in a molecule are additive — a molecule with one double bond has one degree of unsaturation, a molecule with two double bonds has two degrees of unsaturation, and so forth.



6 0
3 years ago
This chart shows the experimental design of four students who are all observing what kind of bird seed birds tend to eat more of
ki77a [65]

Answer:Student 2

Explanation:

Student 2 repeated the experiment several times with different seeds to make sure the experiment would come out with the same answers and was reliable, using the same area would make sure the environment wouldn't interfere. The other students didn't do all the things that student 2 needed for the experiment.

4 0
3 years ago
Read 2 more answers
The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3)
cestrela7 [59]

Answer:

a. The limiting reactant is NaHCO_{3}

b. 0.73 g of carbon dioxide are formed.

c. The grams of excess reactant that do not participate in the reaction are 0333 g.

Explanation:

a)

You know the following reaction:

3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}⇒3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}

First, you determine the molar mass of each compound. For that you must take into account the atomic mass of each element:

  • Na:  23
  • H: 1
  • C: 12
  • O: 16

To determine the molar mass of each compound, you multiply the most atomic of each element present in the molecule by the sub-index that appears after each number, which indicates the present amount of each element in the compound:

  • NaHCO_{3} :23+1+12+16*3=84 g/mol
  • H_{3} C_{6} HO_{7} :1*3+12*6+1*5+16*7= 192 g/mol
  • CO_{2} :12+16*2= 44 g/mol
  • H_{2} O :1*2+16= 18 g/mol
  • Na_{3} C_{6} H_{5} O_{7} : 23*3+12*6+1*5+16*7= 258 g/mol

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of NaHCO_{3} react with 1 mole of H_{3} C_{6} HO_{7}  Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

  • NaHCO_{3} : 252 g
  • H_{3} C_{6} HO_{7} : 192 g

You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:  

If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}

grams of sodium bicarbonate= 1.8375 g

But to perform the experiment you have only 1.4 g of sodium bicarbonate. So <u><em>the limiting reagent is sodium bicarbonate</em></u>.

b)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of NaHCO_{3} react with 3 mole of CO_{2}. Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

  • NaHCO_{3} : 252 g
  • H_{3} C_{6} HO_{7} : 132 g

You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?

grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}

<u><em>grams of carbon dioxide=  0.73 g</em></u>

<u><em>Then, 0.73 g of carbon dioxide are formed.</em></u>

c)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first. This means that citric acid will not react everything, leaving an excess.

To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?

grams of citric acid=\frac{1.4 g * 192 g}{252 g}

grams of citric acid= 1.067 g

But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:

grams of excess reactant=1.4 g - 1.067 g

grams of excess reactant=0.333 g

<em><u>So the grams of excess reactant that do not participate in the reaction are 0333 g.</u></em>

3 0
3 years ago
You place a large pear on a scale and it shows a mass of 200 grams. What is the mass of that pear in centigrams? Remember: King
statuscvo [17]

Answer:

20,000 centigrams

Explanation:

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2 years ago
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