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2 years ago

How many moles of Na2SO4 can be made with 7.28 mol of H2SO4?

Chemistry

1 answer:

Slav-nsk [51]2 years ago

5 0

Answer:

7.28 mol Na2SO4

Explanation:

Since it is already in moles, all we have to do is use a molar ratio

A molar ratio is the proportions of reactants and products using the balanced equation. When writing a mole ratio, the given information must cross out with the right thing.

7.28 mol H2SO4 * 1 mol Na2SO4/1 H2SO4 = 7.28 mol Na2SO4

*notice how the H2SO4 crosses out

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Anestetic [448]

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3 years ago

Be sure to answer all parts. In the average adult male, the residual volume (RV) of the lungs, the volume of air remaining after

Alenkinab [10]

Answer:

For a: The number of moles of air present in the RV is 0.047 moles

For b: The number of molecules of gas is $2.83\times 10^{22}$

Explanation:

For a:

To calculate the number of moles, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of the air = 1.00 atm

V = Volume of the air = 1200 mL = 1.2 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of the air = $37^oC=[37+273]K=310K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of air = ?

Putting values in above equation, we get:

$1.00atm\times 1.2L=n_{air}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 310K\\n_{air}=\frac{1.00\times 1.2}{0.0821\times 310}=0.047mol$

Hence, the number of moles of air present in the RV is 0.047 moles

For b:

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.047 moles of air will contain $(0.047\times 6.022\times 10^{23})=2.83\times 10^{22}$ number of gas molecules.

Hence, the number of molecules of gas is $2.83\times 10^{22}$

7 0

2 years ago

You carefully weigh out 11.00 g of caco3 powder and add it to 44.55 g of hcl solution. you notice bubbles as a reaction takes pl

Zinaida [17]

The bubbles that were observed after the mixing of the two substances is one of the products of the reaction. It is the carbon dioxide that is produced. To determine the mass of this gas produced, we need to remember the Law of conservation of mass where mass cannot be created or destroyed. With this, we can say that the total mass that goes in a process should be equal to the mass that is goes out of the process no matter what the reaction is. We do as follows:

Mass of reactants = mass of products
11.00 + 44.55 = 51.04 + mass of carbon dioxide
mass of carbon dioxide = 4.51 g

8 0

3 years ago

What is the pH for a solution that has an OH ion

Basile [38]

Assuming that you mean 10^-4 M then this would be basic and would have a pH of 10.

pOH = -log[OH].
So pOH = 4
pH=14-pOH
pH = 10

3 0

2 years ago

Is 13.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode?

Andrews [41]

Answer:
Zn =â‡’ Zn+2(0.10) + 2e- (anode)
Zn+2(?M) + 2e- === Zn(s) (cathode)

Zn + Zn+2(?M) ===â‡’ Zn+2(0.10) + Zn
E = E^o -0.0592 log Q; in this case E^o is zero.
E = - 0.0592 /n logQ where n is the number of electrons transferred, in this
case n = 2
23 mV x 1 volt/1000mv = 0.023 Volts
0.023 = -0.0592 / 2 log(0.10) / [Zn+2]
0.023 = -0.0296 { log 0.10 â€“ log [Zn+2] }
0.023 = -0.0296{ -1 - log[Zn+2] }
0.023 = +0.0296 + 0.0296log[Zn+2]
-0.0066 = 0.0296log[Zn+2]
-0.22= log[Zn+2]
[Zn+2] = 10^-0.22 = 0.603 Molar

6 0

3 years ago