Answer:
30.4 g. NH3
Explanation:
This problem tells us that the hydrogen (H2) is the limiting reactant, as there is "an excess of nitrogen." Using stoichiometry (the relationship between the various species of the equation), we can see that for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
But before we can use that relationship to find the number of grams of ammonia produced, we need to convert the given grams of hydrogen into moles:
5.4 g x [1 mol H2/(1.008x2 g.)] = 2.67857 mol H2 (not using significant figures yet; want to be as accurate as possible)
Now, we can use the relationship between H2 and NH3.
2.67857 mol H2 x (2 mol NH3/3 mol H2) = 1.7857 mol NH3
Now, we have the number of moles of ammonia produced, but the answer asks us for grams. Use the molar mass of ammonia to convert.
1.7857 mol NH3 x 17.034 g. NH3/mol NH3 = 30.4 g. NH3 (used a default # of 3 sig figs)
The element is likely to be metal.
The substances that are pure and only consists of atoms are called elements. They are the smallest and the fundamental unit of the compound and cannot be split to form any entity.
<h3>What are metals?</h3>
The metals are the elements that are good conductors of charge, electricity, are opaque, malleable, lustrous and ductile. They have high density and melting points. They differ from the non-metals and metalloids based on their physical and chemical properties.
The metals are more likely to donate and gain electrons to form complexes and can produce transitions. The electron gets excited when the energy is given in the form of heat and electricity and thus when comes to the ground state gives a distinct colouration.
Therefore, the element is likely to be metal.
Learn more about metals here:
brainly.com/question/24911138
Answer:
no it is a heterogeneous mixture because salt and pepper are not mixed uniformly.
Answer:
276g of Br are present
Explanation:
To solve this question we need to find the moles of CaBr2 using its molar mass -CaBr2: 199.89g/mol-. As 1 mole of CaBr2 contains 2 moles of Br we can find the moles of Br and its mass:
<em>Moles CaBr2:</em>
345g * (1mol/199.89g) = 1.7259 moles CaBr2
<em>Moles Br:</em>
1.7259 moles CaBr2 * (2mol Br / 1mol CaBr2) = 3.4519 moles Br
<em>Mass Br -Molar mass: 79.904g/mol-</em>
3.4519 moles Br * (79.904g/mol) = 276g of Br are present