A. <span>The amplitude doubled hope this helps and have a nice day</span>
Answer:
Molecular formula for the gas is: C₄H₁₀
Explanation:
Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g
At STP → 1 atm and 273.15K
1 atm . 0.0336 L = n . 0.082 . 273.15 K
n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)
n = 1.500 × 10⁻³ moles
Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m
Now we propose rules of three:
If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C
58 g of gas (1mol) would have:
(58 g . 0.480) / 0.580 = 48 g of C
(58 g . 0.100) / 0.580 = 10 g of H
48 g of C / 12 g/mol = 4 mol
10 g of H / 1g/mol = 10 moles
Glucose molecule has 6 carbon atoms
Since
potassium and phosphate is what we are to find for and they are both found in
the potassium phosphate solution, therefore we solve for this one first on the
basis of the phosphate.
The formula
for finding the volume given the concentration and number of moles is:
Volume =
number of moles / concentration in Molarity
Volume
potassium phosphate required = 30 mmol phosphate / (3 mmol / mL)
<u>Volume
potassium phosphate required = 10 mL</u>
This would
also contain potassium in amounts of:
Amount of
potassium in potassium phosphate = 10 mL (4.4 meg / mL)
Amount of
potassium in potassium phosphate = 44 meg
Therefore
the potassium chloride required is:
Volume of
potassium chloride = (80 meg – 44 meg) / (2 meg / mL)
<span><u>Volume of
potassium chloride = 72 mL</u></span>
Answer:
Y = 92.5 %
Explanation:
Hello there!
In this case, since the reaction between lead (II) nitrate and potassium bromide is:
Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:
Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:
And the resulting percent yield:
Regards!
