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3 years ago

The energy required to remove an electron from the outermost shell is called

Chemistry

1 answer:

Damm [24]3 years ago

6 0

Answer:

Ionization energy

Explanation:

Ionization energy can be defined as the energy needed to remove an electron from the outermost she'll.

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We have a cylinder. Calculate the following:

Nata [24]

Answer:

mass = 2.65 [nearest hundredth].

Explanation:

Mass = Density / Volume

mass = 1000 / 376.8

mass = 2.653927813163.. ≈ 2.65

8 0

3 years ago

What is the half-life of a radioisotope if a 50-g sample becomes 25 g after 18

vredina [299]

Answer:

18 DAYS IS THE HALF LIFE OF THE RADIOISOTOPE.

Explanation:

Using the formula

Nt = No * (1/2)^t/t1/2

where;

Nt = amount remaining = 25 g

No = Initial amounrt of the radioisotope = 50 g

t = time elapsed = 18 days

t1/2 = half life = unknown

Substitute the values into the equation and obtain the half life of the radioisotope;

Nt = No * (1/2) ^t/t1/2

25 = 50 * (1/2) ^18/t1/2

25 /50 = (1/2)^18/ t1/2

1/2 = (1/2)^18/t1/2

1/2)^1 = (1/2)^18/t1/2

1 = 18/t1/2

t1/2 = 18 days.

So therefore the half life of the radioisotope is 18 days.

4 0

3 years ago

How many moles of al2(so4)3 can form under these conditions?

olchik [2.2K]

A. Since you are starting with equal moles of the two reactants, you can see from the balanced equation that you need 3 moles of H2SO4 for every 2 moles of Al(OH)3. So, H2SO4 is the limiting reactant.

B. 0.420 mol H2SO4 X (1 mol Al2(SO4)3 / 3 mol H2SO4) = 0.140 mol Al2(SO4)3

C. Moles of Al(OH)3 used:

0.420 mol H2SO4 X (2 mol Al(OH)3/3 mol H2SO4)
= 0.280 mol Al(OH)3 used Moles remaining
= 0.420 - 0.280
= 0.140 mol Al(OH)3 remaining.

8 0

4 years ago

(ii) Calculate the maximum mass of ethanol that could be obtained<br> g from 30.0 g of glucose.

Step2247 [10]

Taking into account the reaction stoichiometry, 15.33 grams ethanol could be obtained from 30 grams of glucose.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂

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:

C₆H₁₂O₆: 1 mole
C₂H₅OH: 2 moles
CO₂: 2 moles

The molar mass of the compounds is:

C₆H₁₂O₆: 180 g/mole
C₂H₅OH: 46 g/mole
CO₂: 44 g/mole

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

C₆H₁₂O₆: 1 mole ×180 g/mole= 180 grams
C₂H₅OH: 2 moles ×46 g/mole= 92 grams
CO₂: 2 moles ×44 g/mole= 88 grams

<h3>Mass of C₂H₅OH formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 180 grams of glucose form 92 grams of ethanol, 30 grams of glucose form how much mass of ethanol?

$mass of ethanol=\frac{30 grams of glucosex 92 grams of ethanol}{180 grams of glucose}$