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3 years ago

Ions that are present before and after a neutralization reaction are

Chemistry

1 answer:

VikaD [51]3 years ago

8 0

Answer:

spectator ions

Explanation:

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How many molecules are present in 4.5 moles of H2O?

Gekata [30.6K]

Answer:

Mole of the H2O = 4.5

Number of molecules =4.5 multipled by avogadro's number.

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3 years ago

Zn(CrO4)2 What is the name of the compound

insens350 [35]

Answer:

Zinc chromate

Explanation:

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2 years ago

During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200.

Travka [436]

Answer: The mass of HCl present in 500 mL of acid solution is 36.5 grams

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=200.mL$

Putting values in above equation, we get:

$1\times M_1\times 50.00=1\times 0.500\times 200\\\\M_1=\frac{1\times 0.500\times 200}{1\times 50.00}=2M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Molar mass of HCl = 36.5 g/mol

Molarity of solution = 2 M

Volume of solution = 500 mL

Putting values in above equation, we get:

$2mol/L=\frac{\text{Mass of solute}\times 1000}{36.5g/mol\times 500}\\\\\text{Mass of solute}=\frac{2\times 36.5\times 500}{1000}=36.5g$

Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams

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3 years ago

Calculate the cost of gasoline for a 320 mile trip from Boston to New

Talja [164]

The cost of gasoline for the trip would cost $57.28 because you first divide 320 by 20 which gives you 16 than after that you do 16 multiplied by 3.58 which gives you $57.28 as your final product

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3 years ago

Which carboxylic acid is used to prepare the ester shown?

valentinak56 [21]

Answer:

Explanation:

The general equation for the reaction of a carboxylic acid with an alkanol to form an ester is shown below;

RCOOH + ROH ------> RCOOR + H2O

Hence; the reactant carboxylic acid can only be the compound (CH3)2-CH-CH2-COOH in accordance with the general reaction equation shown above.

Hence the reaction is;

(CH3)2-CH-CH2-COOH + CH3-CH2OH -------> CH3CH2 OCO-CH2-CH-(CH3)2

3 0

2 years ago