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Ahat [919]
3 years ago
10

your friend wants to fly a kite over the water on the beach. which time would you go time a time c or time f

Chemistry
1 answer:
alexandr1967 [171]3 years ago
7 0

Answer:

uh time c

Explanation:

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What is the molality of a solution in which 0.42 moles aluminum chloride has been dissolved in 4200 water ?
madreJ [45]

Answer:

0.1 M

<h3>Explanation:</h3>
  • Molarity refers to the concentration of a solution in moles per liter.
  • It is calculated by dividing the number of moles of solute by the volume of solvent;
  • Molarity = Moles of the solute ÷ Volume of the solvent

<u>In this case, we are given;</u>

  • Number of moles of the solute, NH₄Cl as 0.42 moles
  • Volume of the solvent, water as 4200 mL or 4.2 L

Therefore;

Molarity = 0.42 moles ÷ 4.2 L

            = 0.1 mol/L or 0.1 M

Thus, the molarity of the solution will be 0.1 M

7 0
3 years ago
George Washington Carver received _____ from God during early morning walks in the woods.?
BabaBlast [244]

Answer:

1. Guidance

2. Inspiration

I think this is right hope it helps

8 0
3 years ago
What is the overall ionic equation for zinc + silver nitrate?
3241004551 [841]

Answer:

Zn(NO₃)₂

Explanation:

this single replacement reaction will produce silver metal, Ag , and aqueous zinc nitrate, Zn(NO3)2 . Zinc is above silver is the metal reactivity series, so it will replace silver in silver nitrate

7 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
Fictitious element X has an a average atomic mass of 254.9 and only 2 isotopes, One of its isotopes has an abundance of 72.00 an
elena-14-01-66 [18.8K]

Answer:

265.2amu

Explanation:

Given parameters:

Atomic mass  = 254.9amu

Abundance of isotope 1 = 72%

Atomic mass of isotope 1  = 250.9amu

Abundance of isotope 2  = 100  - 72  = 28%

Unknown:

Atomic mass of isotope 2 = ?

Solution:

To find the atomic mass of isotope 2, use the expression below:

Atomic mass = (abundance of isotope 1 x atomic mass of isotope 1) + (abundance of isotope 2 x atomic mass of isotope 2)

 Now insert the parameters and find the unknown;

  254.9  = (0.72 x 250.9)  + (0.28 x Atomic mass of isotope 2)

 254.9  = 180.648 + 0.28x atomic mass of isotope 2

 254.9 - 180.648  = 0.28x atomic mass of isotope 2

    74.25  = 0.28 x atomic mass of isotope 2

    Atomic mass of isotope 2 = 265.2amu

8 0
3 years ago
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