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Yuki888 [10]
3 years ago
13

What do calcium chloride, sodium carbonate, calcium carbonate, and potassium carbonate all have in common? (other than they reac

t with acid’
Chemistry
2 answers:
EastWind [94]3 years ago
8 0

Answer:

They all have carbonate?

Explanation:

VARVARA [1.3K]3 years ago
7 0

Answer:

the all dissolve in water

Explanation:

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What is the empirical formula for a compound if a sample contains 1.0 g of S and 1.5 g of O?
sammy [17]
I think it’s SO3 I’m not quite sure though
6 0
2 years ago
(a) Calculate the energy released by the alpha decay of 222 86Rn.
Alinara [238K]

Answer:

a). The energy released by the alpha decay of 222 Rn is to 218 Po :

= 5.596 MeV

b).The energy of the alpha particle is 5.5 MeV

c) The recoil energy of Po = 0.096 MeV

Explanation:

The equation for the alpha decay of Rn to Polonium is:

_{86}^{222}\textrm{Rn}\rightarrow _{2}^{4}\textrm{He}+_{84}^{218}\textrm{Po}

The energy of the decay process can be calculated using:

\Delta E=\Delta mc^{2}

\Delta m = Change in the mass

Mass of Po = 218.008965 u

Mass of He = 4.002603 u

Mass of Rn = 222.017576 u

\Delta m = mass of Po + mass of He - mass of Rn

=  4.002603 + 218.008965-222.017576  

= - 0.006008

\Delta E=m(931.5MeV)

\Delta E=-0.006008\times 931.5MeV

= -5.596 MeV

<u><em>The negative sign means energy is released during the process.</em></u>

b) The energy of the alpha particle is :

\frac{Po\ mass }{Rn\ mass}\times E

\frac{218.0089}{222.0175}\times \Delta E

\frac{218.0089}{222.0175}\times 5.596

= 5.494 MeV

= 5.5 MeV

c).

Recoil energy: When the parent nucleus is at rest before the decay then , there must be the some recoil of daughter nucleus to conserve the momentum.This is termed as recoil energy.

<u><em>The energy of the recoil polonium atom :</em></u>

The formula for recoil energy is :

<em>The total energy - the kinetic energy</em>

<em>= 5.596 - 5.5 </em>

<em>= 0.096 MeV</em>

7 0
3 years ago
My Teacher didn't teach us this please help
oksian1 [2.3K]

Answer:

first; why is it a question if teacher never teached you it LOL anyways..

Explanation:

i would say its the second one

"The smae quality of each element is present on both sides of the equation."

5 0
3 years ago
Which is radioactive decay
Liula [17]

Answer:

Explanation:all

7 0
3 years ago
A solution is prepared by dissolving 33.0 milligrams of sodium chloride in 1000. L of water. Assume a final volume of 1000. lite
zheka24 [161]

Answer:

a. Molarity of NaCl solution = 5.64 * 10⁻⁷ mol/L

b. molarity of Na⁺ = 5.64 * 10⁻⁷ mol/L

c. molarity of Cl⁻ = 5.64 * 10⁻⁷ mol/L

d. Osmolarity = 1.128 osmol

e. mass percent of NaCl = 3.30 * 10⁻⁶ %

f. parts per million NaCl = 0.033 ppm NaCl

g. parts per billion of NaCl = 33 ppb of NaCl

h. From the values obtained from e, f and g, the most convenient to use and understand is parts per billion as it has less of a fractional part to deal with especially since the solute concentration is very small.

Explanation:

Molarity of a solution = number of moles of solute (moles)/volume of solution (L)

where number of moles of solute = mass of solute (g)/molar mass of solute (g/mol)

a. Molarity of NaCl:

molar mass of NaCl = 58.5 g/mol, mass of NaCl = 33.0/1000) g = 0.033g

number of moles of NaCl = 0.033/58.5 = 0.000564 moles

Molarity of NaCl solution = 0.000564/1000 = 5.64 * 10⁻⁷ mol/L

b. Equation for the dissociation of NaCl in solution: NaCl ----> Na⁺ + Cl⁻

From the above equation I mole of NaCl dissociates to give 1 mole of Na⁺ ions,

Therefore molarity of Na⁺ = 1 * 5.64 * 10⁻⁷ mol/L = 5.64 * 10⁻⁷ mol/L

c. From the above equation I mole of NaCl dissociates to give 1 mole of Cl⁻ ions,

therefore molarity of Cl⁻ = 1 * 5.64 * 10⁻⁷ mol/L = 5.64 * 10⁻⁷ mol/L

d. From the above equation, dissociation of NaCl in water produces 1 mol Na⁺ and 1 mole Cl⁻.

Total number of particles produced = 2

Osmolarity of solution = number of particles * molarity of siolution

Osmolarity = 2 * 5.64 * 10⁻⁷ mol/L = 1.128 osmol

e. mass of percent of NaCl = {mass of NaCl (g)/ mass of solution (g)} * 100

density of water = 1 Kg/L

mass of water = 1 Kg/L * 1000 L = 1000 kg

1Kg = 1000 g

Therefore mass of solution in g = 1000 * 1000 = 1 * 10⁶ g

mass percent of NaCl = (0.033/1 * 10⁶) * 100 = 3.30 * 10⁻⁶ %

f. Parts per million of NaCl:

parts per million = 1 mg of solute/L of solution

One thousandth of a gram is one milligram and 1000 ml is one liter, so that 1 ppm = 1 mg per liter = mg/Liter.

Since the density of water is 1kg/L = 1,000,000 mg/L

1mg/L = 1mg/1,000,000mg or one part in one million.

parts per million NaCl = 33.0/1000 L = 0.033 ppm NaCl

g. Parts per billion = 1 µg/L of solution

1 g = 1000 µg

therefore, 33.0 mg = 33.0 * 1000 µg = 3.30 * 10⁴ µg

parts per billion of NaCl = 3.30 * 10⁴ µg/1000 L = 33 ppb of NaCl

h. From the values obtained from e, f and g, the most convenient to use and understand is parts per billion as it has less of a fractional part to deal with especially since the solute concentration is very small.

5 0
3 years ago
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