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Lorico [155]
3 years ago
6

A chemistry student is recording the electric charges of some atoms. Use the results in the chart to answer the following questi

ons.
Atom Electric Charge
Carbon −4
Nitrogen +5
Oxygen −2
Part A:
Write an equation to find the difference in electric charge between the Nitrogen atom and the Oxygen atom.

Part B:
Using the equation you wrote in Part A, what is the difference in electric charge between the two atoms?
Chemistry
1 answer:
o-na [289]3 years ago
4 0
Answer Can someone here please look at my questions and see if you can help
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HELP PLEASE!
tia_tia [17]

Vocabulary. Balanced chemical equation: A chemical equation in which the number of each type of atom is equal on the two sides of the equation.

Hope I helped! (´▽`)

__________________________________________________________

単語。平衡化学反応式:各タイプの原子の数が方程式の両側で等しい化学反応式。

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5 0
3 years ago
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If a solution has a pH of 10, is the solution acidic or basic?Explain why
natka813 [3]
Basic because a pH greater than 7 is basic
5 0
3 years ago
If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio
bulgar [2K]

Answer:

m_{Cd(OH)_2}=36.6 gCd(OH)_2

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2}  =0.75molCd(OH)_2

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2

Best regards.

4 0
3 years ago
If you start with 10ml of 0.75 m cu(no3)2 how much cu (s) in grams should be recovered in step #7
LenaWriter [7]
<span>0.48 grams. Not a well worded question since it's assuming I know the reactions. But I'll assume that since there's just 1 atom of copper per molecule of Cu(NO3)2, that the reaction will result in 1 atom of copper per molecule of Cu(NO3)2 used. With that in mind, we will have 0.010 l * 0.75 mol/l = 0.0075 moles of copper produced. To convert the amount in moles, multiply by the atomic weight of copper, which is 63.546 g/mol. So 0.0075 mol * 63.546 g/mol = 0.476595 g. Round the results to 2 significant figures, giving 0.48 grams.</span>
4 0
3 years ago
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