Answer:
1160mg
Explanation:
Molarity = number of moles ÷ volume
According to the information in the question, molarity = 1.25 M, volume = 18.0 mL = 18/1000 = 0.018L
M = n/V
n = M × V
n = 1.25 × 0.018
n = 0.0225moles.
Using mole = mass/molar mass, to find the mass of lithium phosphide (Li3P)
Molar mass of Li3P = 6.9(3) + 31 = 51.7g/mol
mole = mass/molar mass
0.0225 = mass/51.7
mass = 1.16grams.
In milligrams (mg), mass of Li3P = 1.16 × 1000 = 1160mg
Answer:
Continental moderate climet
Answer:
Become familiar with the chemicals to be used, including exposure or spill hazards.
Locate the spill kits and understand how they are used.
Explanation:
Hope this helped! :)
<u>Answer:</u>
<u>For a:</u> The empirical formula of the compound is 
<u>For b:</u> The empirical formula of the compound is 
<u>Explanation:</u>
We are given:
Percentage of P = 43.6 %
Percentage of O = 56.4 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of P = 43.6 g
Mass of O = 56.4 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Phosphorus =
Moles of Oxygen = 
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.406 moles.
For Phosphorus = 
For Oxygen = 
Converting the moles in whole number ratio by multiplying it by '2', we get:
For Phosphorus = 
For Oxygen = 
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of P : O = 2 : 5
Hence, the empirical formula for the given compound is 
We are given:
Percentage of K = 28.7 %
Percentage of H = 1.5 %
Percentage of P = 22.8 %
Percentage of O = 56.4 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of K = 28.7 g
Mass of H = 1.5 g
Mass of P = 43.6 g
Mass of O = 56.4 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Potassium =
Moles of Hydrogen =
Moles of Phosphorus =
Moles of Oxygen = 
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.735 moles.
For Potassium = 
For Hydrogen = 
For Phosphorus = 
For Oxygen = 
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of K : H : P : O = 1 : 2 : 1 : 4
Hence, the empirical formula for the given compound is 