Answer:
4560 Torr
Explanation:
<u>Boyle's Law</u>
where:
- = initial volume
- = final volume
- = initial pressure
- = final pressure
Given:
Substituting the given values into the formula:
To convert atm to Torr, multiply atm by 760:
Answer: The density of glycerine is
Explanation: We are given density of glycerine in which is
To convert t into , we are given conversion rates:
454 grams = 1 pound
So, 1 gram will be equal to
Similarly,
So, will be equal to
Converting the density of glycerin from to using above conversion rates, we get:
Density of glycerin is
A) 0.875 M of MgBr2 is 0.875 mol/L
B) 23.7 mL = 0.0237 L
0.875 mol => 1 L
x mol => 0.0237 L
Cross multiply
1x = 0.875 × 0.0237
x = 0.0207 mol
>> In 23.7 mL of 0.875 M MgBr2 solution there is 0.0207 moles of MgBr2
C) Molar mass of MgBr2 (*) = 24.305 + (2 × 79.904) = 184.113 g/mol
184.113 g => 1 mol
x g => 0.0207 mol
Cross multiply
1x = 184.113 × 0.0207
x = 3.8111 g
>> 0.0207 moles of MgBr2 is equivalent to 3.81 g of MgBr2
>> In 23.7 mL of 0.875 M MgBr2 solution there is 3.81 g of MgBr2
(*) Use your periodic table
Answer:
2.2 x 10²² molecules.
Explanation:
- Firstly, we need to calculate the no. of moles in (6.0 g) sodium phosphate:
<em>no. of moles = mass/molar mass </em>= (6.0 g)/(163.94 g/mol) = <em>0.0366 mol.</em>
- <em>It is known that every mole of a molecule contains Avogadro's number (6.022 x 10²³) of molecules.</em>
<em />
<u><em>using cross multiplication:</em></u>
1.0 mole of sodium phosphate contains → 6.022 x 10²³ molecules.
0.0366 mole of sodium phosphate contains → ??? molecules.
<em>∴ The no. of molecules in 6.0 g of sodium phosphate</em> = (6.022 x 10²³ molecules)(0.0366 mole)/(1.0 mole) = <em>2.2 x 10²² molecules.</em>
The first step is to balance the equation:
<span>C3H8 + 5O2 ---> 3CO2 + 4H2O
Check the balance
element left side right side
C 3 3
H 8 4*2 = 8
O 5*2=10 3*2 + 4 = 10
Then you have the molar ratios:
3 mol C3H8 : 5 mol O2 : 3 mol CO2 : 4 mol H2O
Now you have 40 moles of O2 so you make the proportion:
40.0 mol O2 * [3 mol CO2 / 5 mol O2] = 24.0 mol CO2.
Answer: option D. 24.0 mol CO2
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