1) we can calculate the molecular weight of H₃PO₄
atomic mass (H)=1 amu
atomic mass (P)=31 amu
atomic mass (O)=16 amu
molecular weight (H₃PO₄)=3(1)+31+4(16)=98 amu.
1 mol (H₃PO₄)=98 g
1 mol= 6.022 * 10²³ molecules.
2) we calculate the mass of 4.00*10²³ molecules.
98 g-------------------6.022*10²³ molecules
x------------------------4.00*10²³ molecules
x=(98 g * 4.00*10²³ molecules) / 6.022*10²³ molecules≈65 g
Answer: 65 g
Answer:
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Explanation:
silghjnb bj i bb b nv bv hc U and lamenting produce a few of the tree so much for your time is common to both of us ☺️ to all the said that death of the tree so it cut in cannot be avoided as well as well but the mortal Kombat X the said
Answer: 3.4 atm
Explanation:
Given that:
Volume of gas V = 5L
(since 1 liter = 1dm3
5L = 5dm3)
Temperature T = 0°C
Convert Celsius to Kelvin
(0°C + 273 = 273K)
Pressure P = ?
Number of moles of gas n = 0.75 moles
Note that Molar gas constant R is a constant with a value of 0.0821 atm dm3 K-1 mol-1
Then, apply ideal gas equation
pV = nRT
p x 5dm3 = 0.75 moles x (0.0821 atm dm3 K-1 mol-1 x 273K)
p x 5dm3 = 16.8 atm dm3
p = (16.8 atm dm3 / 5dm3)
p = 3.4 atm
Thus, a pressure of 3.4 atm is exerted by the gas.
Water could be made to boil lower than its normal boiling point of 100 degrees Celsius at 92 degrees Celsius by lowering the atmosphere or external pressure.
The boiling point is the temperature at which the vapor pressure of liquid equals the external or atmosphere pressure.
So, if you decrease the external pressure the temperature needed for the liquid reach the lower external pressure is also lower..
You can accomplish by taking the water to higher levels in the Earth (atmosphere pressure at high altitudes is lower than at sea level) or by creating vaccum.
The given question is incomplete. The complete question is:
Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis. Determine the mass of glucose (C6H1206) produced
Answer: 60.0 g of glucose
Explanation:
To calculate the moles, we use the equation:
a) moles of 
b) moles of 
According to stoichiometry :
6 moles of require = 6 moles of
Thus 2.0 moles of require=
of
Thus is the limiting reagent as it limits the formation of product.
As 6 moles of give = 1 moles of glucose
Thus 2.0 moles of give =
of glucose
Mass of glucose = 
Thus 60.0 g of glucose will be produced from 88.0 g of carbon dioxide and 64.0 g of water
