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harina [27]
2 years ago
9

14. Which of the following groups is most at risk for developing this disease?

Chemistry
1 answer:
ki77a [65]2 years ago
8 0

Answer: c

Explanation: built different

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This element has 80 electrons and is considered a transitional element. What am I?
sdas [7]

Answer:its mercury i think!!!!!

Explanation:

8 0
2 years ago
The electron configuration of an element is 1s^ 2 2s^ 2 2p^ 2 How many valence electrons does the clement have? (1 point )
Monica [59]

Answer:

:)

Explanation:

8 0
2 years ago
Calculate the pH of a buffer solution that contains 0.25 M benzoic acid (C 6H 5CO 2H) and 0.15M sodium benzoate (C
gtnhenbr [62]

Answer:

pH=3.97

Explanation:

Hello,

In this case, for the calculation of the pH of the given buffer we need to use the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa for benzoic acid is 4.19, the concentration of the base is 0.15 M (sodium benzoate) and the concentration of the acid is 0.25 M (benzoic acid), therefore, the pH turns out:

pH=4.19+log(\frac{0.15M}{0.25M} )\\\\pH=3.97

Regards.

4 0
3 years ago
How much volume would a 834.01g pile of sugar have given that it has a density of 1.59g/mL?
boyakko [2]

Answer:

v = 534.5mL

m = 597.15g

Density = 9.23g/mL

Density = 9.125g/mL

Explanation:

Density = mass/ volume

For the first question

Density = 1.59g/mL

Mass = 834.01g

Volume = ?

Using the above formula we have 1.59 = 834.01/v

v = 834.01/1.59

v = 534.5mL

For the second question

Density =0.9167g/mL

Volume = 651.41mL

Mass =?

Using the above formula we have

0.9167 =m/651.41

Cross multiply

m = 0.9167 x 651.41

m = 597.15g

For the third question

Mass =803.44g

Volume=87.03mL

Density =?

Density = 803.44/87.03

= 9.23g/mL

For the fourth

Density = 56.85/6.23

= 9.125g/mL

7 0
3 years ago
Write a net ionic equation for the overall reaction that occurs when aqueous solutions of sodium hydroxide and hydrosulfuric aci
Helga [31]

Answer:

2OH-(aq) + 2H+(aq) → 2H2O(l)

Explanation:

Step 1: Data given

sodium hydroxide = NaOH

hydrosulfuric acid = H2S

Step 2: The unbalanced equation

NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

Step 3: Balancing the equation

On the left side we have 1x Na (in NaOH), on the right side we have 2x Na (in Na2S). To balance the amount of Na on both sides, we have to multiply NaOH on the left side by 2.

2NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

On the left side we have 4x H (2x in NaOH and 2x in H2S), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side by 2. Now the equation is balanced.

2NaOH(aq) + H2S(aq) → Na2S(aq) + 2H2O(l)

Step 4: The net ionic equation

2Na+(aq) + 2OH-(aq) + 2H+(aq) + S^2-(aq) → 2Na+(aq) + S^2-(aq) + 2H2O(l)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2OH-(aq) + 2H+(aq) → 2H2O(l)

8 0
3 years ago
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