N2 + 3H2 --> 2NH3
When 100g of N2 , no of moles of N2= 100/(28)=3.57 mol
no. of moles of h2 = 6/(2)=3mol
Therefore h2 is limiting reagent.
no. of moles of ammonia= 3/3*2=2moles
mass of ammonia produced= 2 mol * (14+3)= 34g
Carbon-12 has the following electron configuration: 1s2-2s2-2p2. As seen in the configuration, the right answer is: the first electron shell has 4 orbitals.
Answer:
84.8%
Explanation:
Step 1: Given data
Bob measured out 1.60 g of Na. He forms NaCl according to the following equation.
Na + 1/2 Cl₂ ⇒ NaCl
According to this equation, he calculates that 1.60 g of sodium should produce 4.07 g of NaCl, which is the theoretical yield. However, he carries out the experiment and only makes 3.45 g of NaCl, which is the real yield.
Step 2: Calculate the percent yield.
We will use the following expression.
%yield = real yield / theoretical yield × 100%
%yield = 3.45 g / 4.07 g × 100% = 84.8%
<em><u />the correct option is <u>c ) 10,000...</u>
an acid which is more acidic than 6 pH will be of pH 5 acid because if one acid is stronger than the other acid than its magnitude of strength is 1000... it is constant always...</em>
Toichiometry time! Remember to look at the equation for your molar ratios in other problems.
31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag 6851.65⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ → ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ = 107.9 g Ag ∅ | 63.5 g Cu | 1 mol Cu | 1 mol Ag 63.5
There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.
