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3 years ago

How is 8.2 x 10^4 - 6.3 x 10^3 written in scientific notation

Chemistry

1 answer:

ser-zykov [4K]3 years ago

6 0

Answer:

= 7.57 × 104

(scientific notation)

= 7.57e4

(scientific e notation)

= 75.7 × 103

(engineering notation)

(thousand; prefix kilo- (k))

Explanation:

Just in case this is all of them

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Draw the alcohol that is the product of the reduction of 3-methylpentanal.

natali 33 [55]

Answer: The product from the reduction reaction is

CH3-CH2-CH(CH3)-CH2-CH2OH

IUPAC name; 3- Methylpentan-1-ol

Explanation:

Since oxidation is simply the addition of oxygen to a compound and reduction is likewise the addition of hydrogen to a compound.

Therefore, hydrogen is added onto the carbon atom adjacent to oxygen in 3- methyl pentanal

CH3 CH2 CHCH3 CH2 CHO thereby -CHO( aldehyde functional group) are reduced to CH2OH ( Primary alcohol) which gives;

3-methylpenta-1-ol .

The structure of the product is:

CH3-CH2-CH(CH3)-CH2-CH2OH

7 0

4 years ago

Find how many milliliters of NaOH should be used to reach the half-equivalence point during the titration of 20.00 mL 0.888 M bu

Varvara68 [4.7K]

<u>Answer:</u> The volume of NaOH solution required to reach the half-equivalence point is 0.09 mL

<u>Explanation:</u>

The chemical equation for the dissociation of butanoic acid follows:

$CH_3CH_2CH_2COOH\rightleftharpoons CH_3CH_2CH_2COO^-+H^+$

The expression of $K_a$ for above equation follows:

$K_a=\frac{[CH_3CH_2CH_2COO^-][H^+]}{[CH_3CH_2CH_2COOH]}$

We are given:

$[CH_3CH_2CH_2COOH]=0.888M\\K_a=1.54\times 10^{-5}$

$[CH_3CH_2CH_2COO^-]=[H^+]$

Putting values in above expression, we get:

$1.54\times 10^{-5}=\frac{[H^+]^2}{0.888}$

$[H^+]=-0.0037,0.0037$

Neglecting the negative value because concentration cannot be negative

To calculate the volume of base, 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 butanoic acid

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

We are given:

$n_1=1\\M_1=\frac{0.0037}{2}M\text{ (half equivalence)}\\V_1=20.00mL\\n_2=1\\M_2=0.425M\\V_2=?mL$

Putting values in above equation, we get:

$1\times \frac{0.0037}{2}\times 20.00=1\times 0.425\times V_2\\\\V_2=\frac{1\times 0.0037\times 20}{1\times 0.425\times 2}=0.087mL$

Hence, the volume of NaOH solution required to reach the half-equivalence point is 0.09 mL

5 0

3 years ago

2. Suppose this was done on top of a balance. Do you think the mass would change as the reaction proceeded?

Viefleur [7K]

Yes because gas is produced

8 0

3 years ago

At 95°C, the pressure of a sample of chlorine is 1.15atm. What will the pressure be at 105°C, assuming constant volume?

Jobisdone [24]

Answer:

95+273=368k

105+273=378k

(1 .15/368)×378

=1.18

4 0

3 years ago

Which pairs of angles in the figure below are vertical angles? Check all that apply.

olga2289 [7]

I think it's option C and option D but if you can only choose one answer then I think it's option C

4 0

3 years ago

Read 2 more answers