A compound of Calcium is used for disinfecting drinking water. Identify the compound.

3. How many atoms in 3.91 moles of sodium?

Gala2k [10]

Answer:

Explanation:

The formula for sodium is Na. It does not form a molecule in some way.

1 mol Na = 6.02*10^23 atoms

3.91 mol = x Cross multiply

x = 3.91 * 6.02 * 10^23

x = 23.65 * 10^23

x = 2.365 * 10^24

Scientific notation is always expressed as a number 1 ≤ x < 10

3 0

3 years ago

A bacteria culture begins with 15 bacteria which double in amount at the end of every hour. Solve for the number of bacteria tha

pshichka [43]

The number of bacteria is given by:
N(t) = N(o) x 2ⁿ
Where N(t) is the number after n hours have passed and N(o) is the original number which is 15.
The number grown in the 12th hour is the difference in the number after the 11th and the 12th hour. Thus:
15 x 2¹² - 15 x 2¹¹
= 30,720 bacteria

5 0

3 years ago

Using spdf notation, what is the electron configuration for a neutral atom of beryllium

kkurt [141]

Answer:

Beryllium

Explanation:

4 0

2 years ago

A solution contains an unknown amount of dissolved magnesium. Addition of

Scrat [10]

Taking into account the reaction stoichiometry, 2.13 grams of magnesium was dissolved in the solution.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Mg²⁺(aq) + Na₂CO₃(aq) → MgCO₃(s) + 2 Na⁺(aq)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Mg²⁺: 1 mole
Na₂CO₃: 1 mole
MgCO₃: 1 mole
Na⁺: 2 moles

The molar mass of the compounds is:

Mg²⁺: 24.3 g/mole
Na₂CO₃: 106 g/mole
MgCO₃: 84.3 g/mole
Na⁺: 23 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Mg²⁺: 1 mole ×24.3 g/mole= 24.3 grams
Na₂CO₃: 1 mole ×106 g/mole= 106 grams
MgCO₃: 1 mole ×84.3 g/mole=84.3 grams
Na⁺: 2 moles ×23 g/mole= 46 grams

<h3>Mass of magnesium dissolved</h3>

The following rule of three can be applied: If by reaction stoichiometry 1 mole of Na₂CO₃ react with 24.3 grams of magnesium, 0.0877 moles of Na₂CO₃ react with how much mass of magnesium?

$mass of magnesium=\frac{0.0877 moles of Na_{2}C O_{3}x24.3 grams of magnesium }{1 mole of Na_{2}C O_{3}}$