Answer:
Explanation:
The <em>final product</em> of the<em> metal</em> and <em>chlorine</em> will contain only the total original amount of molybdenum metal (0.1845 g) and chlorine.
Hence, you can determine the amount of chlorine in the <em>compound</em> by subtraction:
- Amount of chlorine in the compound = amount of compound - amount of molybdenum metal
- Amount of chlorine = 0.5936 g - 0.1845 g = 0.4091 g
The <em>empirical formula</em> is the formula that represents the ratio of atoms in the compound using the least positive integer numbers.
Then, you need to convert the masses in grams to number of moles, for which you use the atomic mass of each element.
<u>1. Atomic masses:</u>
- Atomic mass of molybdenum, Mo: 95.94 g/mol
- Atomic mass of chlorine, Cl: 35.453 g/mol
<u>2. Number of moles:</u>
Number of moles = mass in grams / atomic mass
- Number of moles of Mo: 0.1845 g / 95.94 g/mol = 0.001923 mol
- Number of moles of Cl: 0.4091 g / 35.453 g/mol = 0.01154 mol
<u>3. Ratio of moles:</u>
Divide each number of moles by the least number of moles:
- Mo: 0.001923 / 0.001923 = 1
- Cl: 0.01154 / 0.001924 = 6
Ratio:
Hence, the <em>empirical formula</em> is: 
I think the correct answer from the choices listed above is option C. It would be Cl2 (g) + 2Kl (s) = 2KCl (s) + l2 (g) that <span>describes a single-replacement reaction where Cl replaces I. Hope this answers the question. Have a nice day.</span>
We are given with a compound, Aluminum Hydroxide (Al(OH)3), with a molar of 5.00 x
formula units. We are tasked to solve for it's corresponding mass in g. We need to solve first the molecular weight of Aluminum Hydroxide, that is
Al=27 g/mol
O=16 g/mol
H=1g/mol
Al(OH)3= 27 g/mol +16(3) g/mol +1(3) g/mol= 78 g/mol
Not that 1 mol=6.022x10^{23} formula units, hence,
5.00
formula units x
x
=0.648 g of Al(OH)3
Therefore, the mass of Aluminum Hydroxide is 0.648 g
Explanation:
in the case of blood loss, you need blood from someone with your blood type or with universal donor type
Answer:
= 1.553 x 10²³ ions
Explanation:
One mole of a compound or a substance consists of 6.02 x 10²³ ions. 6.02 x 10²³ is also referred to as Avogadro's number/constant.
From the above question, we have been asked to determine the number of ions in 16g of H₂CO3.
The relative formula mass (RFM) of H₂CO3 is 62.026 u:
Derived as;
H₂CO3 = (2 x 1.008) + (12.01 x 1) + (3 x 16.00) u
= 62.026 u.
Number of moles = Mass/RFM
1 mole = 6.02 x 10²³ ions
0.2580 moles = ?
=
Hence the number of ions in 16g H2CO3 = 1.553 x 10²³
