With that informatio you can:
1) Write the chemical equation
2) Balance the chemical equation
3) State the molar ratios
4) Predict if precipitation occurs.
I will do all four, for you:
1) Chemical equation:
mercury(I) nitrate potassium bromide mercury(I) bromide potassium nitrate
<span>Hg2(NO3)2 + KBr → Hg2Br2 + KNO<span>3
2) Balanced chemical equation
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<span>Hg2(NO3)2 + 2KBr → Hg2Br2 + 2KNO<span>3
3) Molar ratios or proportions:
1 mol </span></span><span>Hg2(NO3)2 : 2 mol KBr : 1 mol Hg2Br2 : 2 mol KNO<span>3
4) Prediction of precipitation.
You can use the solubility rules or a table of solubilities. I found in a table of solutiblities that mercury(I) bromide is insoluble and potassium bromide is soluble, Then you can predict that the precipitation of mercury(I) bromide will occur.
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Answer:
The Cotton Gin
Explanation:
The cotton gin: A game-changing social and economic invention. On this day in 1793, young inventor Eli Whitney had his U.S. patent for the cotton gin approved, an invention that would definitely have an impact on social and economic conditions that led to the Civil War.
Answer: Belongs to the group 2A
Explanation:
As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.
Not only loses easily the first electron, but the second too
To remove the third electron you requiered a huge amount of energy
Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.
The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.
Now the question: group I or group II ?
The elements of group I have low ionization energies for the first electron but high energies for the second ones.
Being all that said, the unknown element belongs to the Group 2A
Answer:
General Formulas and Concepts:
<u>Chemistry - Gas Laws</u>
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
- Charles' Law:
Explanation:
<u>Step 1: Define</u>
Initial Volume: 5.0 L H₂ gas
Initial Temp: 273 K
Final Temp: 985 K
Final Volume: ?
<u>Step 2: Solve for new volume</u>
- Substitute:
- Cross-multiply:
- Multiply:
- Isolate <em>x</em>:
- Rewrite:
<u>Step 3: Check</u>
<em>We are given 2 sig figs as the smallest. Follow sig fig rules and round.</em>
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O.N. of Na = +1
O.N. of O = -2
Let, O.N. of Tin = x
1*2 + x + -2*2 = 0
2+x-4 = 0
x-2 = 0
x = 2
SO OPTION C IS YOUR ANSWER......
