Answer:
14.02 x 10²⁴ molecules
Explanation:
To solve this problem we will use <em>Avogadro's number</em>: It relates to the number of molecules (or atoms) contained in 1 mol of any given substance:
- 14.02 x 10²⁴ molecules ÷ 6.023 x 10²³ molecules/mol = 23.28 moles
Thus, 23.28 moles of water contain 14.02 x 10²⁴ molecules.
Answer:
1.25 moles of HF
Explanation:
1 mole = 20.0 grams of HF
So, to calculate mass of 25.0 grams
=25.0*1/20.0
=1.25 moles
Answer:
The noble gases(neon, helium, argon)
Explanation:
Explanation:
First, let's rewrite the equation here:
4 NH3 + 6 NO -> 5 N2 + 6 H2O.
The equation tells us that 6 moles of NO produces 5 moles of N2.
But the question wants us to find the quantity in grams of NO required if we have 121 g of N2.
1. So first, let's transform 121 g of N2 into moles, using its molar mass (28 g/mol) and the following formula: moles = mass/molar mass
2. Then, we use the equation ratio between NO and N2: 6:5.
3. Then we transform the result into grams of NO, using its molar mass (30.01 g/mol) and the following formula: mass = moles*molar mass
1. moles = 121/28
moles = 4.3 moles of N2
2. 6 moles of NO --- 5 moles of N2
x moles of NO --- 4.3 moles of N2
5x = 4.3*6
x = 25.9/5
x = 5.2 moles of NO
3. mass = 5.2*30.01
mass = 155.6 g
Answer: It will require 155.6 g of NO.
Answer:
C. by its chemical Symbol
Hope this helps!
