Answer:
36g of H2O.
Explanation:
We'll begin by writing the balanced equation for the reaction:
2NaOH + H2SO4 —> Na2SO4 + 2H2O
Next, we shall determine the mass of NaOH that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:
Molar mass of NaOH = 23 + 16 + 1 = 40g/mol
Mass of NaOH from the balanced equation = 2 x 40 = 80g
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H2O from the balanced equation = 2 x 18 = 36g.
From the balanced equation above, we can see evidently that:
80g of NaOH reacted to produce 36g of H2O.
Answer:
It lies in in the <u>visible</u><u> </u><u>region</u><u>.</u>
Frequency: <u>is</u><u> </u><u>7</u><u>.</u><u>4</u><u>0</u><u>7</u><u> </u><u>×</u><u> </u><u>1</u><u>0</u><u>^</u><u>-</u><u>1</u><u>4</u><u> </u><u>Hz</u>

Explanation:
V is speed of light.
f is frequency
lambda is the wavelength
As we can see the chemical equation is balanced.K3PO4 + Al(NO3)3 → 3KNO3 + AlPO4
So, by principle of conservation of mass when 1 mole of K3PO4 reacts with 1 mol of Al(NO3)3 , it peoduces 3 mol of KNO3 and 1 mol of AlPO4
So, when 2.5 moles of potassium phosphate react and Al(NO3)3 is present in excess , 2.5*3= 7.5 mol of KNO3 is formed
Answer: the answer is B!
Explanation:
S and p’s are valence electrons and if added, there is 7 in total !
Answer:
-1.05 V
Explanation:
A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.
Oxidation half equation;
Sn(s) ------> Sn^2+(aq) + 2e
Reduction half equation:
Mn^2+(aq) + 2e ----> Mn(s)
Cell voltage= E°cathode - E°anode
E°cathode= -1.19V
E°anode= -0.14 V
Cell voltage= -1.19 V - (-0.14V)
Cell voltage= -1.05 V