Answer:
The answer to your question is: solvent
Explanation:
Solvent is a component in a solution that is present in the highest amount and is able to dissolve the other or others components.
You need the set of reactions that goes from ammonia to nitric acid.
<span>
1) 4NH3(g)+5O2(g)-->4NO(g)+6H2O(g)
2) 2NO(g)+O2(g)-->2NO2(g)
3) 3NO2(g)+H2O(l)-->2HNO3(aq)+NO(g)
State the ratio of moles of HNO3 to NH3:
4 moles of NH3 produce 4 mole of NO,
4 moles of NO produce 4 moles of NO2
4 moles of NO2 produce 4 * (2 / 3) moles of HNO3 = 8/3 moles of HNO3.
=> (8/3) moles HNO3 : 4 moles NH3
Calculate the number of moles of HNO3 in 900.00 l of 0.140 M solution
M = n / V => n = M * V = 0.140 M * 900.00 liter = 126 moles HNO3
Use proportions:
(</span><span>8/3) moles HNO3 / 4 moles NH3 = 126 moles HNO3 / x
=> x = 126 moles HNO3 * 4 moles NH3 / (8/3 moles HNO3) = 189 moles NH3
Convert moles to grams:
molar mass NH3 = 14 g/mol + 3 * 1g/mol = 17 g/mol
mass in grams = number of moles * molar mass = 189 moles * 17 g/mol = 3213 g
Answer: 3213 g.
</span>
Answer:
The statement true is A. A and B are isotopes of the same element.
Explanation:
Isotopes correspond to atoms of the same element with different mass numbers (A), that is to say they differ in the amount of neutrons in the nucleus. In the case of compound A it has 10 neutrons, and B has 11.
Hello!
If a reaction occurs when a piece of metal is placed in a solution, you can conclude that the solution is <span>probably acidic because bases rarely react with metals.
Strong Acids, like HCl, react with metals to produce salts and release gaseous hydrogen (H</span>₂) which is evidenced by the generation of bubbles in the solution. The general chemical equation for this kind of reactions for a metal M and an acid HA is:
2M(s) + 2HA(aq) → 2MA(aq) + H₂(g)
Have a nice day!
<span>
</span>
The minimum volume of 0.2M sodium sulfide that will precipitate out aluminum from 50.0 mL, 0.25 M aluminum nitrate would be 0.094 L or 94 mL
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
Mole ratio of Na2S and Al(NO3)3 = 3:2
Mole of 50.0 mL, 0.25 M Al(NO3)3 = 50/1000 x 0.25
= 0.0125 mole
Equivalent mole of Na2S = 3/2 x 0.0125
= 0.0188 mole
Volume of 0.2M, 0.0188 mole Na2S = 0.0188/0.2
= 0.094 L or 94 mL
More on stoichiometric calculations can be found here: brainly.com/question/8062886
