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mel-nik [20]
4 years ago
13

I need to find the reactants and products for each of these 7 problems. I need it in 3/5 hours help please!!!

Chemistry
1 answer:
Nezavi [6.7K]4 years ago
4 0

Answer:

The arrow points from the reactants to the products, so just follow the arrows.

Explanation:

some have the reactants on the left and the products on the right, and others are the opposite... just know that

reactants---------> products

or

products<-----------reactants

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Consider the set of isoelectronic atoms and ions a2–, b–, c, d+, and e2+. which arrangement of relative radii is correct?
faust18 [17]
The term isoelectronic atoms means that if the given atoms are neutral, they would have the same number of electrons, which is relative to their sizes. Ions with positive charges are called cations, losing electrons. On the other hand, ions with negative charges are called anions, gaining electrons. The more electrons that the atoms have, the bigger it is in size. Hence, the arrangement of the radii of the atoms would have to be,

      e²⁺, d⁺, c , b⁻, a⁻

The arrangement is from smallest to largest.
7 0
4 years ago
identify the law that explains the following observation nitrogen dioxide can be formed by reacting 14 G of nitrogen with 32 gra
Mrrafil [7]

Answer:

The law of definite proportions. I had the same question for chemistry and this is what they said was right so I got 100%.

Explanation:

8 0
3 years ago
Uranium-238 decays to lead-206 with a half-life of 4.5 x 109 yr. Determine how much uranium-238 decays in milligrams (to three s
Mamont248 [21]

Answer:

2.15 mg of uranium-238 decays

Explanation:

For decay of radioactive nuclide-

                        N=N_{0}.(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

where N is amount of radioactive nuclide after t time, N_{0} is initial amount of radioactive nuclide and t_{\frac{1}{2}} is half life of radioactive nuclide

Here N_{0}=4.60 mg, t=4.1\times 10^{9}yr and t_{\frac{1}{2}}=4.5\times 10^{9}yr

So,N=(4.60mg)\times (\frac{1}{2})^{\frac{4.1\times 10^{9}}{4.5\times 10^{9}}}

so, N = 2.446 mg

mass of uranium-238 decays = (4.60-2.446) mg = 2.15 mg

3 0
3 years ago
Calculate the ph of the solution resulting by mixing 20.0 ml of 0.15 m hcl with 20.0 ml of 0.10 m koh
Aliun [14]

Answer:

1.60.

Explanation:

  • The no. of millimoles of HCl = MV = (0.15 M)(20.0 mL) = 3.0 mmol.
  • The no. of millimoles of KOH = MV = (0.10 M)(20.0 mL) = 2.0 mmol.

<em>Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.</em>

<em></em>

∴ M of remaining HCl [H⁺] remaining = (NV)HCl - (NV)KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.

∵ pH = - log[H⁺]

<em>∴ pH = - log[H⁺] </em>= - log(0.025) = <em>1.602 ≅ 1.60.</em>

5 0
4 years ago
Conclusion Questions:
Lerok [7]
That’s the question?
5 0
3 years ago
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