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Anton [14]
3 years ago
9

( HELP ) ANYONE CAN HELP ME WITH THIS QUESTION PLEASE I REALLY NEED HELP !!!

Chemistry
1 answer:
Kisachek [45]3 years ago
3 0

Answer:

too hard middle school student just need points nothing else

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Pls help I will give brainlist
frez [133]

Answer:

I would say B but i am not sure hope it helps

Explanation:

5 0
3 years ago
The bottom number on each element on the Periodic Table is called
Flura [38]

Answer:

The bottom number on each element of the periodic table are called the 4f series or lanthanoids and 5f or actanoids. They are also called inner transition elements.

7 0
3 years ago
Calculate the molarity of a solution made by adding 45.4 g of nano3 to a flask and dissolving it with water to create a total vo
aleksandr82 [10.1K]
Solutions are made up of two non reacting species called solute and solvent. The amount of solute in solvent is known as concentration of that solute. Concentration is often measured in Molarity. Molarity is the amount of solute dissolved in 1 dm3 of solution. Answer to your question is as follow;

3 0
3 years ago
Read 2 more answers
What happens when an acid and alkali react
Norma-Jean [14]
Acid + alkali > salt + water
the acid and alkali neutralise each other 
5 0
3 years ago
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Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th
Black_prince [1.1K]

Answer:

K_{goal}=1.793*10^{-33}

Explanation:

N2(g)+O2(g)⇌2NO(g), K_1 = 4.10*10^{-31}

N2(g)+2H2(g)⇌N2H4(g), K_2 = 7.40*10^{-26}

2H2O(g)⇌2H2(g)+O2(g), K_3 = 1.06*10^{-10}

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g)                     Eq (1)

Equilibrium constant for Eq (1) is K_1*K_3*K_3

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g)                       Eq (2)  

Equilibrium constant for Eq (2) is (K_1*K_3*K_3)^{1/2}

Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}

7 0
3 years ago
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