I need help can you help

irakobra [83]

Answer:

gold

Explanation:

to work out density you do mass which is 38.6 divided by volume which is 2cm cubed and you get the answer of 19.3 so it is gold.

7 0

2 years ago

What is the product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes?

zheka24 [161]

The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .

Sodium borohydride is a relatively selective reducing agent Ethanolic solutions of Sodium borohydride reduces aldehyde , and ketone , in the presence of acid chloride , ester , epoxide , lactones , acids , nitriles , nitro groups.

The sodium borohydride does not reduce ester group because sodium borohydride is not strong enough and the electrophilicity at carbony carbon of ester is not more as compare toaldehyde , and ketone

The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .

to learn more about sodium borohydride and ethanol click here ,

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4 0

2 years ago

he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c

Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer: The rate constant at 324°C is $61.29M^{-1}s^{-1}$

Explanation:

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

$\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_{244^oC}$ = equilibrium constant at 244°C = $6.7M^{-1}s^{-1}$

$K_{324^oC}$ = equilibrium constant at 324°C = ?

$E_a$ = Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $244^oC=[273+244]K=517K$

$T_2$ = final temperature = $324^oC=[273+324]K=597K$

Putting values in above equation, we get:

$\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}$

Hence, the rate constant at 324°C is $61.29M^{-1}s^{-1}$

8 0

4 years ago

25.0 mL of a hydrofluoric acid solution of unknown concentration is titrated with 0.200 M NaOH. After 20.0 mL of the base soluti

lesantik [10]

Answer:

[HF]₀ = 0.125M

Explanation:

NaOH + HF => NaF + H₂O

Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.

=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.

HF ⇄ H⁺ + F⁻

C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)

Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka

[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M

6 0

3 years ago

Exactly how much time must elapse before 16 grams of potassium-42decays, leaving 2 grams of the original isotope?(1) 8 × 12.4 ho

AleksandrR [38]

The answer is (3) 3 × 12.4 hours

To calculate this, we will use two equations:
$(1/2) ^{n} =x$
$t_{1/2} = \frac{t}{n}$
where:
n - number of half-lives
x - remained amount of the sample, in decimals
 $t_{1/2}$ - half-life length
t - total time elapsed.

First, we have to calculate x and n. x is remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams × 100 percent ÷ 16 grams
x = 12.5% = 0.125

Thus:
 $(1/2) ^{n} =x$
 $(0.5) ^{n} =0.125$
$n*log(0.5)=log(0.125)$
$n= \frac{log(0.5)}{log(0.125)}$
$n=3$

It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
 $t_{1/2} = 12.4$
 $t_{1/2} *n = t$
 $t= 12.4*3$

Therefore, it must elapse 3 × 12.4 hours before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope

7 0

3 years ago

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