<h2>
Answer:</h2>
<h3>From the equation it is evident that 2 moles of Sodium metal produces 1 mole of hydrogen gas.</h3><h3>At STP 1mole of any gas occupies a volume of 22.4 litres.</h3><h3>Therefore, 7.80 gives---(7.80x1)/22.4 moles = 0.3482 moles</h3><h3>Since the mole ratio of Sodium to hydrogen is 2:1, then the number of moles of sodium that reacted is given by the following expression.</h3><h3>(0.3482 * 2) / 1 moles which gives 0.6964 moles.</h3><h3>The atomic mass of sodium is 23 thus the mass of sodium that reacted is given by:</h3><h3>mass=no. of Monogram</h3><h3>0.6964 * 23 = 16.02 grams.</h3><h2>
Explanation:</h2><h3>please mark me brainlist</h3>
Explanation:
Make an observation
ask a question
form a hyposisese
make a prediction
test the prediction
use the results to fix the prediction and hypotheses
The formation of ammonia gas involves reacting hydrogen gas and nitrogen gas in a mole ratio of 3 to 1. as shown below:
<h3>What is the equation of the formation of ammonia?</h3>
Ammonia gas is formed from the reaction between nitrogen gas and hydrogen gas.
Three moles of hydrogen gas will react with 1 mole of nitrogen gas to form 2 moles of ammonia gas.
The equation of the reaction is given below as:

Therefore, the formation of ammonia gas involves reacting hydrogen gas and nitrogen gas in a mole ratio of 3 to 1.
Learn more about ammonia gas at: brainly.com/question/7982628
D is the correct answer
every other option contains an element
Explanation:
1.
Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)
2.
Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)
A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.
3.
Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)
2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)
Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)
2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)
4.
The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.
Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.