What does wadding do?

Assume an object A with a mass of 66.789 g and a volume of 10.1 mL. what is its density?

docker41 [41]

Answer: 6.61

Explanation:

5 0

3 years ago

If the gas in the piston above has a volume of 20.0 L at a temperature of 25 C what is the volume of that gas when it is heated

dexar [7]

Answer:

15.98 L

Explanation:

First, you need to find T1, T2, V1 and V2.

T1 = 25 C = 298.15 K (25C + 273.15K)

T2 = 100 C = 373.15 K (100C + 273.15K)

V1 = 20. L

V2 = ? (we are trying to find)

Next, rearrange to fit the formula

V2 = V1 x T1 / T2

Next, fill in with our numbers

V2 = 20. L x 298.15 K / 373.15 K

Do the math and you should get...

15.98 L

- If you need more help or futher explanation please let me know. I would be glad to help!

8 0

3 years ago

Read 2 more answers

Having a child will change your financial obligations, and can lead to a career change.

MrRa [10]

Yes, that's a true statement?

3 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 the limiting reagent is sodium bicarbonate.

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?

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

grams of carbon dioxide= 0.73 g

Then, 0.73 g of carbon dioxide are formed.

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

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

3 0

3 years ago

What is the final volume of a gas (in liters) when the initial volume is 14.00 L at

8_murik_8 [283]

Answer:

$V_2=1344L$

Explanation:

Hello there!

In this case, since we have a problem about volume-pressure relationship, the idea here is to use the Boyle's law to calculate the final volume as shown below:

$P_2V_2=P_1V_1\\\\V_2=\frac{P_2V_2}{P_1}\\$

Then, we plug in the initial and final pressures and the initial volume to obtain:

$V_2=\frac{14.00L*0.9600atm}{0.01000atm}\\\\V_2=1344L$

Regards!

4 0

3 years ago