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olganol [36]
3 years ago
6

Which ion has smaller size and why?Mg++ or Na+.​

Chemistry
1 answer:
alexgriva [62]3 years ago
7 0

Answer:

Explain your answer. mg2+ would be the smaller ion this is because each ion has the same number of electrons however mg2+ has a greater number of protons and therefore is more charge dense and the outer electrons feel a greater pull from the nucleus.

Hope this helps, have a great day/night, and stay safe!

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20. What is the area of a piece of metal foil that measures 43.9 cm by 29.21 cm? Express the answer to
Aneli [31]

Answer:

The correct answer is b. 1280 cm^2

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Kendall wants to determine if there is a trend in air temperature changes during April . Which procedure should she chose?
Alexeev081 [22]
I think the answer is A
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3 years ago
Read 2 more answers
Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose
Slav-nsk [51]

Answer:

The minimum mass of octane that could be left over is 43.0 grams

Explanation:

Step 1: Data given

Mass of octane = 73.0 grams

Mass of oxygen = 105.0 grams

Molar mass octane = 114.23 g/mol

Molar mass oxygen = 32.0 g/mol

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate the number of moles

Moles = mass / molar mass

Moles octane = 73.0 grams / 114.23 g/mol

Moles octane = 0.639 moles

Moles O2 = 105.0 grams / 32.0 g/mol

Moles O2 = 3.28 moles

Step 4: Calculate the limiting reactant

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

O2 is the limiting reactant. It will completely be consumed. (3.28 moles). There will react 3.28 / 12.5 = 0.2624 moles. There will remain 0.639 - 0.2624  = 0.3766 moles octane

Step 5: Calculate mass octane remaining

Mass octane = moles * molar mass

Mass octane = 0.3766 moles * 114.23 g/mol

Mass octane = 43.0 grams

The minimum mass of octane that could be left over is 43.0 grams

3 0
3 years ago
Hurry please!
Pani-rosa [81]

Answer : The mass of of water present in the jar is, 298.79 g

Solution : Given,

Mass of barium nitrate = 27 g

The solubility of barium nitrate at 20^oC is 9.02 gram per 100 ml of water.

As, 9.02 gram of barium nitrate present in 100 ml of water

So, 27 gram of barium nitrate present in \frac{27g}{9.02g}\times 100ml=299.33ml of water

The volume of water is 299.33 ml.

As we know that the density of water at 20^oC is 0.9982 g/ml

Now we have to calculate the mass of water.

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}

\text{Mass of water}=(0.9982g/ml)\times (299.33ml)=298.79g

Therefore, the mass of of water present in the jar is, 298.79 g

5 0
3 years ago
Plz help me solve this question is it A,B,C or D
Rashid [163]

Answer:

B - To increase the rate of the reaction

Explanation:

Catalysts speed up the reaction without being reactants or products, so aren't used up in the reaction.

6 0
2 years ago
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