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Andreas93 [3]
2 years ago
15

Name the salt produced if sodium carbonate reacts with dilute nitric acid ? Write the equation.

Chemistry
1 answer:
Angelina_Jolie [31]2 years ago
5 0

Answer:

2HNO3 +Na2CO3 → CO2 + H2O + 2NaNO3

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Find the number of cations present in 7g of sodium prosphate Na=23 g/mol O=16g/mol P=31g/mol​
Snowcat [4.5K]

Answer:    This contains magnesium, Mg2+, and hydroxide, OH–

, ions. Each magnesium ion is +2 and

each hydroxide ion is -1: two -1 ions are needed for one +2 ion, and the formula for magnesium

hydroxide is Mg(OH)2. The (OH)2 indicates there are two OH–

ions. In a formula unit of

Mg(OH)2, there are one magnesium ion and two hydroxide ions; or one magnesium, two

oxygen, and two hydrogen atoms. The subscript multiplies everything in ( )

hope that helped!!

3 0
3 years ago
I have to circle the one that fits correctly
nadya68 [22]
The answer is chloroplast
8 0
2 years ago
Read 2 more answers
Can someone answer this?
yulyashka [42]

No se si aun necesitas ayuda o no

3 0
2 years ago
For the reaction 2NOCl(g) <-----> 2NO(g) + Cl2(g), Kc = 8.0 at a certain temperature. What concentration of NOCl must be p
Nat2105 [25]
Answer is: <span>concentration of NOCl is 3.52 M.
</span>
Balanced chemical reaction: 2NOCl(g) ⇄ 2NO(g) + Cl₂<span>(g).
Kc = 8.0.
</span>[NOCl] = 1.00 M; equilibrium concentration.
[NO] = x.
[Cl₂] = x/2; equilibrium concentration of chlorine.<span>
Kc = </span>[Cl₂] ·[NO]² / [NOCl].
8.00 = x/2 · x² / 1.
x³/2 = 8.
x = ∛16.
x = 2.52 M.
co(NOCl) = [NOCl] + x.
co(NOCl) = 1.00 M + 2.52 M.
co(NOCl) = 3.52 M; the initial concentration of NOCl.


7 0
2 years ago
(4) Calculate the % of a compound that can be removed from liquid phase 1 by using ONE to FOUR extractions with a liquid phase 2
maksim [4K]

Answer:

One extraction: 50%

Two extractions: 75%

Three extractions: 87.5%

Four extractions: 93.75%

Explanation:

The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.

qⁿ = (V₁/(V₁ + KV₂))ⁿ

We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:

qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ =  (V₁/(2V₁))ⁿ = (1/2)ⁿ

When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.

When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.

When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.

When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.

5 0
2 years ago
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