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jeyben [28]
2 years ago
14

I need help plsss

Chemistry
1 answer:
AveGali [126]2 years ago
4 0

Answer:

Carbon contains four electrons in its outer shell. Therefore, it can form four covalent bonds with other atoms or molecules.

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The ka values for several weak acids are given below. which acid (and its conjugate base) would be the best buffer at ph = 8.0?
Pie
One of the best buffer choice for pH = 8.0 is Tris with Ka value of  6.3 x 10^-9.

To support this answer, we first calculate for the pKa value as the negative logarithm of the Ka value: 
     pKa = -log Ka

For Tris, which is an abbreviation for 2-Amino-2-hydroxymethyl-propane-1,3 -diol and has a Ka value of 6.3 x 10^-9, the pKa is
     pKa = -log Ka
            = -log (6.3x10^-9)
            = 8.2

We know that buffers work best when pH is equal to pKa:
     pKa = 8.2 = pH 

Therefore Tris would be a best buffer at pH = 8.0.
7 0
3 years ago
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Why is gunshot residue important for a forensic scientist? What are the potential complications with this type of evidence?
erik [133]
A residue from a gunshot is most likely gun powder, which tells you what kind of bullet was shot and the type of gun that was used to shoot the target/victim/person. Some complications may be that there is more than one gun or weapon which uses that residue, so it may be hard to pinpoint it and the bullet can't really tell you who it is unless there's DNA on the bullet, and the chemicals of the bullet may even destroy evidence.
3 0
3 years ago
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For two variables in a direct proportion, what is the result of doubling one variable?
Airida [17]

Answer:

the other variable is also doubled

Explanation:

direct proportion, same thing has to happen to both variables

6 0
3 years ago
A rectangular block has the following dimensions: 3.21 dm, 5.83 cm, and 1.84 in. The block has a mass of 1.94 kg. What is the de
spin [16.1K]

The density of the rectangular block in g/mL is 7.0.

<u>Given the following data:</u>

  • Mass of block = 22.8 gra1.94 kg
  • Length of block = 3.21 cm
  • Width of block = 5.83 cm
  • Height of block = 1.84 in.

To find the density of the block in g/mL:

First of all, we would determine the volume of the rectangular block by using the following formula:

Volume = length × width × height

<u>Conversion:</u>

1 in = 2.54 cm​

5.83 in = X cm

Cross-multiplying, we have:

X = 2.54(5.83)\\\\X = 14.81 \; cm

Volume = 3.21 × 5.83 × 14.81

Volume = 277.16 cubic centimeters.

<u>Note</u>: Milliliter (mL) is the same as cubic centimeters.

1000 grams = 1 kg

Y grams = 1.94 kg

Cross-multiplying, we have:

Y = 1940 grams

Now, we can find the density:

Density = \frac{Mass}{Volume}\\\\Density = \frac{1940}{277.16}

<em>Density </em><em>= 7</em><em>.0 g/mL</em>

Therefore, the density of the rectangular block in g/mL is 7.0.

Read more: brainly.com/question/18320053

4 0
3 years ago
g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete
Mandarinka [93]

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899        0.01714        0           0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714    0.01714 (end)

0.00185  | 0 → 0.01714    0.01714 (end)  

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

8 0
3 years ago
Read 2 more answers
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