Answer:
ΔG° of reaction = -47.3 x J/mol
Explanation:
As we can see, we have been a particular reaction and Energy values as well.
ΔG° of reaction = -30.5 kJ/mol
Temperature = 37°C.
And we have to calculat the ΔG° of reaction in the biological cell which contains ATP, ADP and HPO4-2:
The first step is to calculate the equilibrium constant for the reaction:
Equilibrium Constant K =
And we have values given for these quantities in the biological cell:
[HP04-2] = 2.1 x M
[ATP] = 1.2 x M
[ADP] = 8.4 x M
Let's plug in these values in the above equation for equilibrium constant:
K =
K = 1.47 x M
Now, we have to calculate the ΔG° of reaction for the biological cell:
But first we have to convert the temperature in Kelvin scale.
Temp = 37°C
Temp = 37 + 273
Temp = 310 K
ΔG° of reaction = (-30.5 ) + (8.314)x (310K)xln(0.00147)
Where 8.314 = value of Gas Constant
ΔG° of reaction = (-30.5 x ) + (-16810.68)
ΔG° of reaction = -47.3 x J/mol
Answer:
The equation for molarity is moles/liter for the first question you would do 0.256/0.143 liters to get 1.790 mol/L
Explanation:
The second problem you would do need to find the moles of NaCl which you would do by doing 4.89 g/58.44g/mol= 0.08367 then do 0.08367/0.600= 0.139 mol/L
The third problem would be the same steps as the second one.
The fourth problem would be (0.460M)(5.50L)= 2.53 moles
Answer:
Density = 19.3 g/cm³
Explanation:
D = m/v
m = 965 g
v = 50 cm³
so,
965 g / 50 cm³ = 19.3g/cm³
Hope this helps!
<span>the balanced equation for the reaction is as follows ;
Na</span>₂S + 2AgNO₃ ---> 2NaNO₃ + Ag₂<span>S
stoichiometry of Na</span>₂S to AgNO₃<span> is 1:2
number of AgNO</span>₃<span> moles reacted - 0.315 mol/L x 0.04000L = 0.0126 mol according to molar ratio of 1:2
number of Na</span>₂S moles required are - 1/2 x number of AgNO3 moles reacted Na₂<span>S moles = 0.0126 mol /2 = 0.00630 mol
molarity of Na</span>₂<span>S - 0.260 M
there are 0.260 mol in 1 L
therefore 0.00630 mol are in - 0.00630 mol / 0.260 mol/L
volume of Na</span>₂<span>S required = 0.0242 L
volume of Na</span>₂S required = 24.4 mL
Answer:
13.5 mL of solution.
Explanation:
Given:
the solution is of 2% concentration.
The dosage required is 10mg/kg
The weight of dog is 27 kg
Solution:
The mass of barbiturate required for the 27 kg dog = 27 X 10 mg = 270 mg
As the solution is 2%, so there is 2g of barbiturate is dissolved in 100mL of solution
The mass of barbiturate required = 0.270 g
For 2 g we will need 100mL of solution
For 1 g we will need = 50mL of solution
for 0.270g we will need = 50 X 0.270 mL of solution = 13.5 mL of solution.