a) f2 loses electrons and is oxidised. li gains electrons and is reduced
c)sn2+ loses electrons and is oxidised.Al gains electrons and is reduced
b)br2 loses electrons and is oxidised. I- in KI gains electrons and is reduced
Answer:
30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.
Explanation:
M1 = 3.00 M
M2 = 1.80 M
V2 = 50 .0 mL = 50 /1000 L = 0.05 L
V1 = unknown
In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.
So we use the equation:
M1 V1 = M2 V2
V1 = M2 V2 / M1
V2 = 1.80 * 0.05 / 3.0
V2 = 0.09 /3.0
V2 = 0.03 L or 30 mL
To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.
If a gas has an initial pressure of 24,650 pa and an initial volume of 376 ml, then the final volume would be 11,943.8144 ml if the pressure of the gas is changed to 775 torr assuming that the amount and the temperature of the gas remain constant.
It is given that the initial pressure P₁ is 24,650Pa and initial volumeV₁ is 376ml and the final pressureP₂ is 775 torr. We need to find the final volume of the gas. The final volume could be found using the following formula:
P₁V₁ = P₂V₂
By substituting the values, we get
24650 x 376 = 776 x V₂
9268400 = 776V₂
V₂ = 9268400/776
V₂ = 11,943.8144 ml
Therefore, the final volume of the gas would be 11,943.8144 ml
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Answer:
2 Na + 1 Cl2 -> 2 NaCl
Explanation:
The answer is really simple, because if you have 1 nonmetal element that has a subscript of 2, you need to multiply the product and the first reactant by 2 to balance it.
