Answer:
Ok:
Explanation:
So, you can use the Henderson-Hasselbalch equation for this:
pH = pKa + log(
) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.
We can solve that
1 = log(
) and so 10 = or 10HA = A-. For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.
0.216g of aluminium compound X react with an excess of water water to produce gas. this gas burn completely in O2 to form H2O and 108cm^3of CO2 only . the volume of CO2 was measured at room temperature and pressure
0.108 / n = 24 / 1
n = 0.0045 mole ( CO2 >>0.0045 mole
0.216 - 0.0045 = 0.2115
so Al = 0.2115 / 27 => 0.0078 mole
C = 0.0045 * 1000 => 4.5 and Al = 0.0078 * 1000 = 7.8
First, we are using the ideal gas law to get n the number of moles:
PV = nRT
when P is the pressure = 748 mmHg/760 = 0.984 atm
V is the volume = 4 L
R is ideal gas constant = 0.0821
T is the temperature in Kelvin = 300 K
∴ n = 0.984atm*4L/0.0821*300
= 0.1598 moles
when the concentration = moles * (1000g / mass)
= 0.1598 * (1000g / 58 g )
= 2.755 M
when the freezing point = 5.5 °C
and Kf = - 5.12 °C/m
∴ the freezing point for the solution = 5.5 °C + (Kf*m)
= 5.5 °C - (5.12°C/m * 2.755m)
= -8.6 °C
Using the Equation: PV=nRT
Where P is the pressure 60 cmHg or 600 mmHg or 600/760= 0.789 atm
V is the volume 125 ml or 0.125 L, n is the number of moles, R is a constant 0.082057, and T is temperature 25 °C or 298 K;
Therefore:
0.789 × 0.125 = n × 0.082057 × 298
n = 0.0987/24.45
= 0.004036 mol
0.004036 mole has a mass of 0.286 g
Hence; 1 mole has a mass of 0.286/0.004036
= 70.8 g /mol
Therefore the molar mass of the gas is 71 g/mol (2 sfg)
Answer:
64.0
Explanation:
2Mg+O2 ---> 2MgO
use dimentional analysis to find the amount of moles of O2 needed first
4.00molMg x 1.00mol O2/ 2.00 mol Mg=. 2.00 mol O2
using the coefficients you can see the mole ratio for O2:Mg the mole ratio is 1:2 which is why there is 1 mole on the top for 2 moles on the bottom. The Mg would cancel and multiply 4 by 1 then divide by 2, or multipy 4 by 1/2
Now that you have the moles of O2 you use the molar mass to find the grams in 2 moles of O2
2.00 mol O2 x 32.0g/1.00 mol = 64.0 g
multiply 2 by 32
