Answer:
Explanation:
The first step is the <u>calculation of the moles</u> of and , so:
Now, in 1 mol of CO2 we have 1 mol of C and in 1 mol of we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to <u>calculate the grams of C and O</u> and then do the <u>substraction</u> form the initial amount, so:
Now we can <u>convert the grams</u> of O to moles, so:
The next step is to divide all the mol values by the <u>smallest one</u>:
Therefore the formula is
Answer:
Volume of water it can hold 63000,000 cm³
Mass of water is 63000,000 g.
Explanation:
Given data:
Length of tank = 70.00 cm
Width of tank = 300.00 mm
Height of tank = 300.00 m
Volume of water it can hold = ?
Mass of water = ?
Solution:
First of all we will convert the units into same unit.
Width of tank = 300.00 mm × 1 cm / 10 mm = 30 cm
Height of tank = 300.00 m× 100 cm / 1 m = 30000 cm
Volume of water it can hold:
V = l×w×h
V = 70.00 cm × 30 cm × 30000 cm
V = 63000,000 cm³
Mass of water:
Density of water from literature = 1 g/cm³
d = m/v
1 g/cm³ = m/ 63000,000 cm³
m = 63000,000 g
Because it is breaking out into small particals and there is change in shape not composition.
Answer:
amount of silver chloride required is 0.015 moles or 2.1504 g
Explanation:
0.1M AgCL means 0.1mol/dm³ or 0.1mol/L
1L = 1000mL
if 0.1mol of AgCl is contained in 1000mL of solution
then x will be contained in 150mL of solution
cross multiply to find x
x = (0.1*150)/1000
x= 0.015 moles
moles of silver chloride present in 150 mL of solution is 0.15 moles
To convert this to grams, simply multiply this value by the molar mass of silver chloride
molar mass of silver chloride AgCl =107.86 + 35.5
=143.36 g/mol
mass of AgCl = moles *molar mass
=0.015*143.36
=2.1504g
=