Albert Einstein received his first novel prize for his discovery of the law of the photoelectric effect.
Answer:
When chlorine water is added to potassium bromide solution, chlorine will displace the bromine from the salt solution. The chlorine will be REDUCED (gain electrons) to chlorine ions. The bromide ions will be OXIDISED (lose electrons) to become bromine. The bromine will turn the solution orange.
Explanation:
<em><u>H</u></em><em><u>A</u></em><em><u>V</u></em><em><u>E</u></em><em><u> </u></em><em><u>A</u></em><em><u> </u></em><em><u>G</u></em><em><u>O</u></em><em><u>O</u></em><em><u>D</u></em><em><u> </u></em><em><u>D</u></em><em><u>A</u></em><em><u>Y</u></em>
I still do not know but they do have quanza muscles
Answer:
Answer: The correct answer is Option B.
Explanation:
To calculate the number of moles, we use the equation:
\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}Number of moles=Molar massGiven mass ....(1)
For N_2N2 :
Given mass of nitrogen gas = 10 g
Molar mass of nitrogen gas = 28 g/mol
Putting values in above equation, we get:
\text{Moles of iron oxide}=\frac{10g}{28g/mol}=0.357molMoles of iron oxide=28g/mol10g=0.357mol
The given chemical reaction follows:
N_2+O_2\rightarrow 2NON2+O2→2NO
As, oxygen gas is present in excess. Thus, it is considered as an excess reagent and nitrogen is considered as a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
Answer:
120g
Explanation:
Step 1:
We'll begin by writing the balanced equation for the reaction.
Sn + 2HF —> SnF2 + H2
Step 2:
Determination of the number of mole HF needed to react with 3 moles of Sn.
From the balanced equation above,
1 mole of Sn and reacted with 2 moles of HF.
Therefore, 3 moles Sn will react with = 3 x 2 = 6 moles of HF.
Step 3:
Conversion of 6 moles of HF to grams.
Number of mole HF = 6 moles
Molar Mass of HF = 1 + 19 = 20g/mol
Mass of HF =..?
Mass = number of mole x molar Mass
Mass of HF = 6 x 20
Mass of HF = 120g
Therefore, 120g of HF is needed to react with 3 moles of Sn.