You want to divide by avagadros number (6.22 x 10^23). This will cancel the atoms unit and give moles, moles of Iridium. Now you want to calculate the atomic mass of Iridium which is in units of grams per mole. Multiply these two numbers and the moles will cancel giving you grams.
Setting up a dimension analysis type of thing helps tremendously. See what you have to cancel in order to get what you want. We canceled the atoms, then we canceled the moles, and then we got grams.
Answer:
Dalton law states that the partial pressure of a gas in a mixture of gases is equal to the total pressure of the mixture.
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)
Atomic mass Cu = 63.546 a.m.u
63.546 g ---------------- 6.02x10²³ atoms
22 g --------------------- ??
22 x (6.02x10²³ ) / 63.546 => 2.08x10²³ atoms
hope this helps!
Answer: 0.0508mL
Explanation: Using the basic formula that states: C acid * V acid = C base * V base. we have:0.568 * 17.88 = 20 * C base.
therefore concentration of the base is 1.0156/20 = 0.0508 mL
