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3 years ago

Compare and contrast atomic number and atomic mass

Chemistry

1 answer:

salantis [7]3 years ago

5 0

Answer:

The major difference between atomic number and mass number is that the atomic number states the number of protons present in an atom whereas, the mass number indicates the total of the number of protons and the number neutrons present in an atom.

Explanation:

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A chemical ________ involves one or more reactants changing into new products.

guapka [62]

Your answer is C. Reaction

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3 years ago

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Determine the concentrations of BaBr2, Ba2 , and Br– in a solution prepared by dissolving 1.18 × 10–4 g BaBr2 in 1.00 L of water

Anika [276]

Supposing complete ionization:
BaBr2 → Ba{2+} + 2 Br{-} 

(2.23 × 10^–4 g BaBr2) / (297.135 g BaBr2/mol) / (2.00 L) = 3.75 × 10^-7 mol/L BaBr2 

(3.75 × 10^-7 mol/L BaBr2) x (1 mol Ba{2+} / 1 mol BaBr2) = 3.75 × 10^-7 mol/L Ba{2+} 

(3.75 × 10^-7 mol/L BaBr2) x (2 mol Br(-} / 1 mol BaBr2) = 7.50 × 10^-7 mol/L Br{-}

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3 years ago

PLEASE HELP THIS IS DUE IN AN HOUR!!!

Soloha48 [4]

Explanation:

The hydronium ion concentration can be found from the pH by the reverse of the mathematical operation employed to find the pH. [H3O+] = 10-pH or [H3O+] = antilog (- pH) Example: What is the hydronium ion concentration in a solution that has a pH of 8.34? On a calculator, calculate 10-8.34, or "inverse" log ( - 8.34).

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3 years ago

If 1.50 L of 0.780 mol/L sodium sulfide is mixed with 1.00 L of a 3.31 mol/L lead(II) nitrate solution, what mass of precipitate

NARA [144]

Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

Na2S(aq) + Pb(NO3)2(aq) ----> 2NaNO3(aq) + PbS(s)

Ionically;

Pb^2+(aq) + S^2-(aq) -----> PbS(s)

Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide

Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles

Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles

Then we determine the limiting reactant. The limiting reactant yields the least amount of product.

Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide

1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate

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3 years ago

state the reactants and products of this reaction using this chemical reaction: CH2COOH + NaHCO2 YEILDS CH2COONa + H2O +CO2

attashe74 [19]

Am i too late or do you still need the answer? if so i got you

8 0

3 years ago