Answer:
3 p.m most probably.
Explanation:
They found it at Monday 5 pm so its done before that time so its 3 p.m
Answer:
42.2 moles of H3PO4
Explanation:
The equation of the reaction is:
P2O5(s) + 3 H2O(l) ⟶ 2 H3PO4.
First we must obtain the number of moles of P2O5 from
Number of moles of P2O5= reacting mass of P2O5/molar mass of P2O5
Molar mass of P2O5= 141.9445 g/mol
Number of moles= 3000g/141.9445 g/mol = 21.1 moles of P2O5
From the reaction equation;
1 mole of P2O5 yields 2 moles of H3PO4
21.1 moles of P2O5 will yield 21.1 ×2/ 1 = 42.2 moles of H3PO4
Answer: voltage drops in each resistor ΔU= RI
Explanation: if lamps or other resistor which cause load are in series in
Electric circuit, current I passing circuit is same. Voltage decreases
In every resistor
Answer:
The answer you selected is correct (A)have
Explanation:
Red blood cells one goal which is to carry oxygen throughtout your body.. without it you would die to lack of oxygen.
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
![\left[I_2_{Equilibrium} \right]](https://tex.z-dn.net/?f=%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D)
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
