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Nookie1986 [14]
3 years ago
11

CaSO4.1/2 H2O molecular mass​

Chemistry
1 answer:
7nadin3 [17]3 years ago
3 0

I don't care but one another time and I have any questions about this email

You might be interested in
Use the unbalanced equation NH3+O2=NO+H2O to find the mole ratio between NO and H2O
Leona [35]

Answer:

For every 4 moles of NO created, 6 moles of H2O are created so the ratio is 4:6

Explanation:

You just need to balance the equation.

NH3 + O2 -> NO + H2O

1. I started with hydrogen; there's 3 on the left and 2 on the right. Multiply them together to find a number they both go into (3×2=6, but in this case 6 hydrogen on each side does not work so I doubled it so there is 12 hydrogen on each side).

This will bring you to this:

4NH3 + O2 -> NO + 6H2O

2. Now get equal amounts of nitrogen on each side. There's 4 nitrogen on the left side, and 1 on the right. Multiply the right by 4. Then you will have this:

4NH3 + O2 -> 4NO + 6H2O

3. Last thing you need to do is have the same amount of oxygen on both sides. On the left you have 2 and on the right you have 10. Get the left to 10 by multiplying it by 5.

Balanced: 4NH3 + 5O2 -> 4NO + 6H2O

In word form, for every reaction between 4 moles of ammonia and 5 moles of oxygen, 4 moles of nitric oxide and 6 moles of water will be created.

I hope this helps!

5 0
3 years ago
Як визначити амфотерний, луг чи основний елемент в хімії. Коли назви даєш, то пишеш наприклад CuSO4 - амф, купрум сульфат. Чому
Leno4ka [110]
What? Please write in Spanish or English
7 0
4 years ago
Four cars travel at the same speed down a road. Their mass and kinetic energy are shown in the graph,
Alex777 [14]

Answer:B

Explanation:

3 0
2 years ago
How many moles of NaOH are in 27.8 mL of 0.168 M NaOH?
Alex17521 [72]
Your answer would be 0.00285 moles.
8 0
3 years ago
What mass of CO was used up in the reaction with an excess of oxygen gas if 24.7g of carbon dioxide is formed? 2 CO + O2 > 2
Zolol [24]
Balance Chemical Equation,
                                      2 CO  +  O₂   →   2CO₂
Acc. to this reaction,
88 g (2 mole) of CO₂ was produced when  =  56 g (2 mole)of CO was reacted
So, 
24.7 g of CO₂ will be produced by reacting =  X g of CO

Solving for X,
                                    X  =  (56 g × 24.7 g) ÷ 88 g

                                    X  =  2.26 g ÷ 88 g

                                    X  =  0.0257 g of CO

Result:           
            0.0257 g of CO is required to be reacted with excess of O₂ to produce 24.7 g of CO₂.
7 0
3 years ago
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