Answer:
98.8
Explanation:
CsF + XeF6 --> CsXeF7
37.8g ................. ?g
37.8g CsF x (1 mol CsF / 151.9g CsF) x (1 mol CsXeF7 / 1 mol CsF) x (397.2g CsXeF7 / 1 mol CsXeF7) = 98.8g CsXeF7 .......... to three significant digits
Ooooh boy alright. So, this may or may not be a limited reactant problem so we need to first find out of it is.
First, how many moles of each substance are there
the molar mass of BCl3 is <span>117.17 grams so 37.5 g / 117.17 is ~ .32 mol.
The molar mass of H2O is 18.02 so 60 / 18.02 is ~ 3.33 mol.
Now, for every 1 mole of BCl3, there are 3 moles of HCl created. Therefore, BCl3 can create ~ .96 moles.
For every 3 moles of H2O, there are 3 moles of HCl created. Therefore, HCl can create ~3.33 moles.
But, there is not enough BCl3 to support that 3.33 moles, only enough for .96 moles, therefore BCl3 is the limiting reactant. Now, to answer the question, simply multiply .96 moles by the molar mass of HCl.
.96 x 36.46 = ~35 g</span>
Answer:
A = 1,13x10¹⁰
Ea = 16,7 kJ/mol
Explanation:
Using Arrhenius law:
ln k = -Ea/R × 1/T + ln(A)
You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).
Using the values you will obtain:
y = -2006,9 x +23,147
As R = 8,314472x10⁻³ kJ/molK:
-Ea/8,314472x10⁻³ kJ/molK = -2006,9 K⁻¹
<em>Ea = 16,7 kJ/mol</em>
Pre-exponential factor is:
ln A = 23,147
A = e^23,147
<em>A = 1,13x10¹⁰</em>
<em></em>
I hope it helps!
Answer:
True.
Explanation:
This process is known as chemiosmosis in which there is a movement of ions across semipermeable membrane. Hydrogen ions moves from region of its higher concentration to the region of lower concentration. As this process belongs to the diffusion or osmosis of water molecules across cell membrane that is why known as chemiosmosis.
ATP synthase is an enzyme which function is to form ATP by using free energy generated in result of movement of hydrogen ions.
Answer: [H3O+] > [OH-] and Kw = 1 x 10 -14
Explanation:
i just had this question and tried to look it up and i got it right by guessing my bad but i hope this is ur answer as well
