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3 years ago

Calculate the percentage composition of Mg3 (Po4)2

1 answer:

7 0

Mg3(PO4)2 - the molar mass would be 262g/mol, which is 100%

Atomic mass of Mg is 24, since we have 3Mg we multiply by 3 and get a mass of 72

262 : 100% = 72 : x%

x = 72*100 / 262

x = 27.5%

And do that for every element — get the molar mass of P and multiply by 2, use a ratio, and get the molar mass of O and multiply by 8 and use ratios :)

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What is happening when an oxygen molecule is formed from two separate oxygen atoms?

Pepsi [2]

They share four electrons, two from each oxygen atom

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2 years ago

How do hydrocarbons differ in structure from carbohydrates.

ICE Princess25 [194]

Hydrocarbons are carbon and hydrogen. Methane is CH4, and propane is C3H8. Methene is CH3, and propene is C3H6. Carbohydrates are hydrates of carbon. They have the general formula (CH2O)x. thats how it differs

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3 years ago

A student adds sugar to a cup of iced tea and a cup of hot tea. She notices that the time needed for the sugar to dissolve in ea

Natali [406]

Explanation:

the experiment conducted is the student adds sugar to a cup of iced tea and a cup of hot tea. She notices that the time needed for the sugar to dissolve in each cup is different. She thinks this has something to do with the temperature of the tea

hypothesis: If the student puts the sugar in both glasses of tea, then the sugar in the hot tea should dissolve quicker.

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3 years ago

Why does electronegativity increase across a period?

stiv31 [10]

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3 years ago

Suppose 0.795 g of sodium iodide is dissolved in 100. mL of a 39.0 m M aqueous solution of silver nitrate.

lubasha [3.4K]

Answer:

The final molarity of iodide anion is 0.053 M

Explanation:

Step 1: Data given

Mass of sodium iodide (NaI) = 0.795 grams

Volume of the solution = 100 mL = 0.1 L

Molarity of aqueous solution of silver nitrate (AgNO3) = 39 mM = 0.039M

The molecular mass of sodium iodide is 149.89 g/mol.

Step 2: The balanced equation

AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq)

Step 3: Calculate number of moles of sodium iodide

Moles NaI = mass NaI / Molar mass NaI

Moles NaI = 0.795 grams / 149.89 g/mol

Moles NaI = 0.0053 moles

For 1 mole AgNO3 consumed, we need 1 mole NaI to produce 1 mole AgI and 1 mole NaNO3

The sodium iodide will dissociate as followed:

NaI(aq) → Na+(aq) + I-(aq)

Step 4: Calculate iodide ions

For 1 mole NaI, we have 1 mole of I-

For 0.0053 moles of NaI we'll have 0.0053 moles I-

Step 5: Calculate molarity of iodide ion

Molarity = moles I- / volume

Molarity I- = 0.0053 moles / 0.1 L

Molarity I- = 0.053 M

The final molarity of iodide anion is 0.053 M

5 0

3 years ago