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QveST [7]
3 years ago
10

To what volume in millimeters must 50.0 mL of 18.0 M H2SO4 be diluted to obtain 4.35 M H2SO4?

Chemistry
1 answer:
poizon [28]3 years ago
7 0
We know that to relate solutions of with the factors of molarity and volume, we can use the equation: M_{1}  V_{1} = M_{2}  V_{2}

**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.

So now we can assign values to these variables. Let us say that the 18 M H_{2}  SO_{4} is the left side of the equation. Then we have:

(18 M)(0.050 L)=(4.35M) V_{2}

We can then solve for V_{2}:

V_{2}= \frac{(18M)(0.05L)}{4.35M} and V_{2} =0.21 L or 210 mL

We now know that the total amount of volume of the 4.35 M solution will be 210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.


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Which of the following best defines velocity?
Lina20 [59]

Answer: D

Explanation:

Because velocity is speed

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3 years ago
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Vanadium (II) oxide with Iron (III) oxide results in Vanadium (V) oxide and Iron (II) oxide. Balance this equation
evablogger [386]

Answer:

2VO + 3Fe2O3 —> V2O5 + 6FeO

Explanation:

The skeletal equation for the reaction is given below below:

VO + Fe2O3 —> V2O5 + FeO

We can balance the equation above by doing the following:

There are 2 atoms of V on the right side and 1 atom on the left side. It can be balance by putting 2 in front of VO as shown below:

2VO + Fe2O3 —> V2O5 + FeO

Now, we have a total of 5 atoms of O on the left and 6 atoms on the right side. We can balance it by putting 3 in front of Fe2O3 and 6 in front of FeO as shown below:

2VO + 3Fe2O3 —> V2O5 + 6FeO

Now, we can see that the equation is balanced

4 0
2 years ago
A sample of gas at 1.10 atm has a volume of 326 mL. What is the new volume if the pressure is changed to 1.90 atm?
Brut [27]

Answer

For this we use ideal gas equation which is:

P1V1 = P2V2

P1 = 1.10 atm

V1 = 326 ml

P2 = 1.90

V2 = ?

By rearranging the ideal gas equation:

V2 =  P1V1 ÷ P2

V2 = 1.10 × 326 ÷1.90

V2 = 358.6 ÷ 1.90

V2 = 188.7 ml

8 0
3 years ago
Fe2O3+2Al=Al2O3+2Fe
True [87]
We determine the limiting reactant by using the moles present in the equation and the actual moles.
According to equation, ratio of Fe₂O₃ : Al = 1 : 2
Actual moles of Fe₂O₃ = 187.3 / (56 x 2 + 16 x 3)
= 1.17
Actual moles of Al = 94.51 / 27
= 3.5
Fe₂O₃ is limiting. Fe₂O₃ required:
(moles Al)/2 = 3.5/2 = 1.75
Moles to be added = 1.75 - 1.17
= 0.58
Mass to be added = moles x Mr
= 0.58 x (56 x 2 + 16 x 3)
= 92.8 grams
6 0
2 years ago
Read 2 more answers
Freon−12 (CF2Cl2), widely used as a refrigerant and aerosol propellant, is a dangerous air pollutant. In the troposphere, it tra
nevsk [136]

Answer:

141g of CCl₄

Explanation:

First, we have to write the balanced equation.

CCl₄(g) + 2 HF(g) ⇄ CF₂Cl₂(g) + 2 HCl(g)

We can calculate how many moles of CF₂Cl₂ using the ideal gas equation.

V = 14.9 dm³ = 14.9 L

T = 21°C + 273.15 = 294.15 K

P = 1.48 atm

R = 0.08206 atm.L/mol.K

P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1.48atm.14.9L}{0.08206\frac{atm.L}{mol.K}.294.15K }=0.914mol

We can use proportions to find the mass of CCl₄ required to obtain 0.914 moles of CF₂Cl₂. According to the balanced equation, 1 mol of CF₂Cl₂ is produced when 1 mol of CCl₄ reacts. And the molar mass of CCl₄ is 154 g/mol.

0.914molCF_{2}Cl_{2}.\frac{1molCCl_{4}}{1molCF_{2}Cl_{2}} .\frac{154gCCl_{4}}{1molCCl_{4}} =141gCCl_{4}

7 0
3 years ago
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