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2 years ago

What is the mass of a 2.5 mole sample of MgO2

Chemistry

1 answer:

MrRa [10]2 years ago

6 0

Mass = no. of moles x molecular weight
m = n x Mr
m = 2.5 mol x (24 + [16 x 2])
m = 140g

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What is the net ionic equation for a reaction between HCl and NaOH

kodGreya [7K]

Answer:

What is the net ionic equation for a reaction between HCl and NaOH?

Explanation:

A salt is a neutral ionic compound. Let's see how a neutralization reaction produces both water and a salt, using as an example the reaction between solutions of hydrochloric acid and sodium hydroxide. The overall equation for this reaction is: NaOH + HCl → H2O and NaCl

Hope that helped.

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2 years ago

A sailor on a trans-Pacific solo voyage notices one day that if he puts 735.mL of fresh water into a plastic cup weighing 25.0g,

Gennadij [26K]

Answer:

Amount of salt in 1 L seawater = 34 g

Explanation:

According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater

mass of freshwater = density * volume

1 cm³ = 1 mL

mass of freshwater = 0.999 g/cm³ * 735 cm³ = 734.265 g

mass of freshwater + cup = 734.265 + 25 = 759.265 g

Therefore, mass of equal volume of seawater = 759.265 g

Volume of seawater displaced = 735 mL = 0.735 L (assuming the cup volume is negligible)

1 liter = 1000 cm³ = 1000 mL;

Density of seawater = mass / volume

Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L

Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L

mass of 1 Liter seawater = 1033.01 g

mass of 1 Liter freshwater = 999 g

mass of salt dissolved in 1 L of seawater = 1033.01 g - 999 g = 34.01 g

Therefore, amount of salt in 1 L seawater = 34 g

4 0

3 years ago

can someone solve me some papers I don't have time and I need to submit today please I will make you brilliant

dem82 [27]

Answer:

submit the papers onto this platform and we will try to help out.

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3 years ago

Can you help me out plz and thank u

Nonamiya [84]

Answer:

stirring

Explanation:

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3 years ago

Consider the reaction. mc014-1.jpg How many grams of methane should be burned in an excess of oxygen at STP to obtain 5.6 L of c

garri49 [273]

The reaction for the combustion of methane can be expressed as follows.
CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
moles of CO2 = (5.6 L) / (22.4 L/1 mole)
moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
= 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane.

5 0

3 years ago

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