Express the answer in scientific notation and with the correct number of significant figures: (6.32 x 10-4) ÷ 12.64

Chemistry

2 answers:

anygoal [31]3 years ago

6 0

Answer:

$5.00*10^{-5}$

Explanation:

$\frac{6.32*10^{-4} }{12.64} = 0.00005= 5.00*10^{-5}$

The answer is in three significant figures because that is the least number of significant figures in any of the numbers involved in the calculation

elixir [45]3 years ago

4 0

Express the answer in scientific notation and with the correct number of significant figures:
(6.32 x 10-4) ÷ 12.64
5.00 x 10^-5

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A student uses a solution of 1.2 molar sodium hydroxide (NaOH) to calculate the concentration of a solution of sulfuric acid (H2

Mila [183]

From the calculations, the concentration of the acid is 0.24 M.

<h3>What is neutralization?</h3>

The term neutralization has to do with a reaction in which an acid and a base react to form salt and water only.

We have to use the formula;

CAVA/CBVB = NA/NB

CAVANB =CBVBNA

The equation of the reaction is; 2NaOH + H2SO4 ----> Na2SO4 + 2H2O

CA = ?

CB = 1.2 M

VA = 50 mL

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CA = 0.24 M

Learn more about neutralization:brainly.com/question/27891712

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1 year ago

Draw the product that valine forms when it reacts with excess CH3CH2OH and HCl followed by a wash with aqueous base.

-BARSIC- [3]

Answer:

Product: ethyl L-valinate

Explanation:

If we want to understand what it is the molecule produced we have to analyze the reagents. We have valine an amino acid, in this kind of compounds we have an amine group ( $NH_2$ ) and a carboxylic acid group ( $COOH$ ). Additionally, we have an alcohol ( $CH_3CH_2OH$ ) in the presence of HCl (a strong acid) in the first step, and a base ( $OH^-$ ).

When we have an acid and an alcohol in a vessel we will have an esterification reaction. In other words, an ester is produced. As the first step, the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the second step, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In step 3, a proton is transferred to produce a better leaving group ( $H_2O$ ). In step 4, a water molecule leaves the main structure to produce again the double bond C=O. Finally, a base ( $OH^-$ ) removes the hydrogen from the C=O bond to produce ethyl L-valinate