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Ede4ka [16]
3 years ago
9

Why is the addition of cold distilled water to the container holding the aspirin you are trying to isolate a necessary step and

why should this step be carried out carefully?.
Chemistry
1 answer:
Rudik [331]3 years ago
3 0

Answer:

The aspirin is more soluble in ethanol than water - the water helps the crystals separate from solution.

Explanation:

Hope this helps!

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PLEASE HELP ME!!!! Which energy graph represents the nonspontaneous transition of graphite into diamond?
Viefleur [7K]

Answer:

A picture .There is high reactions in progression still the energy reaction reduced to indicate complete reaction.

4 0
3 years ago
NEED ANSWER ASAP!<br> How many grams are in 1.50 moles of bromine liquid?
padilas [110]

Answer:

119.85 grams Br or 120. grams Br (sig figs)

Explanation:

1.50 moles Br          79.90 g Br

--------------------- x ------------------------ = 119.85 grams Br or 120 grams Br (sig figs)

                                     1 mole

7 0
3 years ago
PLS HELP IF YOU WANT A BRAINLIEST BEST ANSWER CHOICE GETS BRAINLIEST Which describes the two cells formed through mitosis?
luda_lava [24]

Answer:C

Explanation:

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6 0
3 years ago
Question 15 How many grams of NaCl are required to make 500.0 mL of a 1.500 M solution? 58.40 g 175.3 g 14.60 g 43.83 g
ExtremeBDS [4]
Hi!

To make 500 mL of a 1,500 M solution of NaCl you'll require 43,83 g

To calculate that, you will need to use a conversion factor to go from the volume of the 1,500 M solution to the required grams. For this conversion factor, you'll use the definition for Molar concentration (M=mol/L) and the molar mass of NaCl. The conversion factor is shown below:

gNaCl=500mLsol* \frac{1L}{1000 mL}* \frac{1,500 mol NaCl}{1Lsol}* \frac{58,4428 g NaCl}{1 mol NaCl} \\ =43,83gNaCl

Have a nice day!
4 0
3 years ago
Determine the pH of 0.050 M HCN solution. HCN is a weak acid with a Ka equal to 4.9 x 10-10<br> DONE
nadezda [96]

Answer:

The pH of the solution is 5.31.

Explanation:

Let "\alpha is the dissociation of weak acid - HCN.

The dissociation reaction of HCN is as follows.

                  HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}

Initial                  C                         0            0

Equilibrium        c(1- \alpha)              c\alpha c\alpha

Dissociation constant = Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}

=\frac{c\alpha^{2}}{(1-\alpha)}

In this case weak acids \alpha is very small so, (1-\alpha ) is taken as 1.

Ka=C\alpha^{2}

\alpha=\sqrt\frac{ka}{c}

From the given the concentration = 0.050 M

Substitute the given value.

\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}

[H_{3}O^{+}]=c\alpha

[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}

pH= -log[H_{3}O^{+}]

=-log[4.9\times10^{-6}]

=6-log 4.9= 5.31

Therefore, The pH of the solution is 5.31.

7 0
3 years ago
Read 2 more answers
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