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Feliz [49]
3 years ago
10

Magnesium sulfide + nitric acid net ionic

Chemistry
1 answer:
Oxana [17]3 years ago
8 0

Answer:

MgSO_3(s) + 2 HNO_3(aq) rightarrow Mg^2+(aq) + 2 NO_3^-(aq) + SO_2(g) + H_2O(l) b.

Explanation:

The balanced net ionic equation for the reaction of magnesium sulfite with nitric acid is a. MgSO_3(s) + 2 HNO_3(aq) rightarrow Mg^2+(aq) + 2 NO_3^-(aq) + SO_2(g) + H_2O(l) b.

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How many total atoms are there in 69.1 g of ammonia (NH3)?​
lozanna [386]

Answer:

answers is - 3.568

I think this correct answer

5 0
3 years ago
If a gas has a volume of 750 mL at 25oC, what would the volume of the gas be at 55oC?
blagie [28]

Answer:

The answer to your question is     V2 = 825.5 ml

Explanation:

Data

Volume 1 = 750 ml

Temperature 1 = 25°C

Volume 2= ?

Temperature 2 = 55°C

Process

Use the Charles' law to solve this problem

                V1/T1 = V2/T2

-Solve for V2

                V2 = V1T2 / T1

-Convert temperature to °K

T1 = 25 + 273 = 298°K

T2 = 55 + 273 = 328°K

-Substitution

                V2 = (750 x 328) / 298

-Simplification

                V2 = 246000 / 298

-Result

                V2 = 825.5 ml

             

7 0
3 years ago
How much water should be added to 85 mL of 0.45 M HCI to reduce the concentration to 0.20 M?
valkas [14]

Answer:

106.25 mL

Explanation:

For this, we can use

C1×V1=C2×V2

C1 = 0.45

V1 = 85

C2= 0.20

V2= ?

0.45 × 85 = 0.20 × V2

V2= (0.45 × 85)/0.20

V2=191.25mL

To find the amount of water added, subtract V1 from V2

191.25 - 85 =106.25mL

8 0
4 years ago
A mixture of CO2 and Kr weighs 35.0 g and exerts a pressure of 0.708 atm in its container. Since Kr is expensive, you wish to re
ollegr [7]

Explanation:

Since, it is given that carbon dioxide is completely removed by absorption with NaOH. And, pressure inside the container is 0.250 atm.

For Kr = 0.250 atm and pressure CO_{2} will be calculated as follows.

           CO_{2} = (0.708 - 0.250) atm

                       = 0.458 atm

Now, we will calculate the mole fraction as follows.

            CO_{2} = \frac{0.458}{0.708}

                       = 0.646

               Kr = \frac{0.250}{0.708}

                    = 0.353

Now, we will convert into gram fraction as follows.

              CO_{2} = 0.646 \times 44

                          = 28.424

                   Kr = 0.353 \times 83.78

                        = 29.57

Therefore, total mass is calculated as follows.

           Total mass = (28.424 + 29.57)

                              = 57.994

Hence, the percentage of CO_{2} and Kr are calculated as follows.

          CO_{2} = \frac{28.424}{57.99} \times 100

                     = 49%

               Kr = \frac{29.57}{57.99} \times 100

                    = 51%

Hence, amount of CO_{2} and Kr present i mixture is as follows.  

         CO_{2} in mixture = 35 \times 0.49

                             = 17.15 g

                  Kr = 35 \times 0.51

                       = 17.85 g

Thus, we can conclude that 17.15 g of CO_{2} is originally present and 17.85 g of Kr is recovered.

6 0
3 years ago
Select the correct answer.
Svetradugi [14.3K]

Answer:

the answer is A

Explanation:

According to the law of conservation of energy or the first law thermodynamics energy neither be created nor destroyed, energy is transferred from one form to another form.

4 0
3 years ago
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