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3 years ago

Choose which of the following are organic. (check all that apply) *

Chemistry

1 answer:

Charra [1.4K]3 years ago

7 0

Answer:

The answer is A, B and E

Explanation:

A - carbohydrates

B - Lipids

C- COH

E - Proteins

all except O and NaOH

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2 years ago

1. A 1500kg car is being lifted by a hydraulic jack attached to a flat plate. Underneath the plate is

AURORKA [14]

Answer:

the pressure in the pipe in the case when there is no net force on the car is 81,726 N/m^2

Explanation:

a. The computation of the pressure in the pipe in the case when there is no net force on the car is shown below

As we know that

Pressure = F ÷ area

Also

F = mg

Now

= (1500 × 9.8) ÷π (0.24)^2

= 81,726 N/m^2

Hence, the pressure in the pipe in the case when there is no net force on the car is 81,726 N/m^2

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3 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$

calculating the reverse reaction

$H_2(g) + I_2(g)\rightleftharpoons 2HI(g)$

$Kc = \frac{1}{Kc} \\\\$

$= \frac{1}{0.034386}\\ \\= 29.081\\$

7 0

3 years ago