All categories

3 years ago

Can someone please help me, I’m taking a test

Chemistry

1 answer:

lawyer [7]3 years ago

8 0

Answer:

Im pretty sure its none of these

You might be interested in

A 3.45 microgram sample of Uranium has a mass of how many grams?

Ivan

There are 3.45e-6 grams.

4 0

3 years ago

Conduction, convection and radiation can occur in variety of ways. Give another example, like the campfire picture above, where

Pavlova-9 [17]

Answer:

Explanation:

It shows the answer in the text.

4 0

3 years ago

Given M1V1 = M2V2

Radda [10]

4.37*v=2.33*0.755
V=1.76/4.37
V=0.4025 l

4 0

3 years ago

What could you predict about a methane molecule that would most likely be made by using both the quantum model of the atom and J

Svetlanka [38]

By using the quantum model of the atom and John Dalton's model you can predict that the methane molecule will most likely be made up of Hydrogen and Carbon.

5 0

3 years ago

A chemistry graduate student is given 125.mL of a 0.20M acetic acid HCH3CO2 solution. Acetic acid is a weak acid with =Ka×1.810−

Over [174]

Answer : The mass of sodium acetate is, 1.097 grams.

Explanation : Given,

The dissociation constant for acetic acid = $K_a=1.8\times 10^{-5}$

Concentration of acetic acid (weak acid)= 0.20 M

volume of solution = 125. mL

pH = 4.47

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (1.8\times 10^{-5})$

$pK_a=5-\log (1.8)$

$pK_a=4.74$

Now we have to calculate the concentration of sodium acetate (conjugate base or salt).

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

Now put all the given values in this expression, we get:

$4.47=4.74+\log (\frac{[Salt]}{0.20})$

$[Salt]=0.107M$

Now we have to calculate the mass of sodium acetate.

$\text{Concentration}=\frac{\text{Mass of }NaCH_3CO_2\times 1000}{\text{Molar mass of }NaCH_3CO_2\times \text{Volume of solution (in mL)}}$

$0.107M=\frac{\text{Mass of }NaCH_3CO_2\times 1000}{82g/mol\times 125mL}$

$\text{Mass of }NaCH_3CO_2=1.097g$

Therefore, the mass of sodium acetate is, 1.097 grams.

6 0

4 years ago