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Vinvika [58]
2 years ago
6

Mass of NaOH = 0.2 Volume of solution 500 ml

Chemistry
1 answer:
Yuliya22 [10]2 years ago
6 0
Density will be mass/volume= 0.2/500 = 0.0004
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Please draw the resonance structure for this compound indicating formal charges and lone pairs
ikadub [295]

Here are the resonance contributors I found.

8 0
3 years ago
Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe
leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

c = \frac{n}{V}

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}

M_{KOH} = 56.106 \frac{g}{mol}

<u>Calculate the moles of potassium hydroxide.</u>

n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}

n_{KOH} = 0.25941(9)mol

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

V = \frac{500.0mL}{1}*\frac{1L}{1000mL}

V = 0.5000L

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

c = \frac{n}{V}

c = \frac{0.25941(9)mol}{0.5000 L}              

c = 0.5188(3) \frac{mol}{L}         <= Keep an insignificant figure for rounding

c = 0.5188 \frac{mol}{L}              <= Rounded up

c = 0.5188M               <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0
3 years ago
Consider the reaction at 500 ° C 500°C . N 2 ( g ) + 3 H 2 ( g ) − ⇀ ↽ − 2 NH 3 ( g ) K c = 0.061 N2(g)+3H2(g)↽−−⇀2NH3(g)Kc=0.06
Sav [38]

Answer:

Q = 0.061 = Kc

Explanation:

Step 1: Data given

Temperature = 500 °C

Kc=0.061

1.14 mol/L  N2

5.52 mol/L H2

3.42 mol/L NH3

Step 2: Calculate Q

Q=[products]/[reactants]=[NH3]²/ [N2][H2]³

If Qc=Kc then the reaction is at equilibrium.  

If Qc<Kc then the reaction will shift right to reach equilibrium.

If Qc>Kc then the reaction will shift left to reach equilibrium.  

Q = (3.42)² / (1.14 * 5.52³)

Q = 11.6964/191.744

Q = 0.061

Q = Kc the reaction is at equilibrium.  

4 0
3 years ago
Which aldehyde is an intermediate in the reduction of ethyl benzoate with lithium aluminum hydride?
garri49 [273]

Answer:

tetrahedral aldehyde

Explanation:

  1. The reaction begins with a hydride nucleophile reacting with the ester carbonyl carbon to form the tetrahedral intermediate.
  1. The carbonyl reforms to produce an aldehyde with the loss of the alkoxide ion.
  2. The resulting aldehyde undergoes a subsequent reaction with a hydride nucleophile to form another tetrahedral intermediate. The carbonyl is not able to reform, because there are no stable leaving groups.
  3. Therefore, the tetrahedral intermediate is protonated to produce a primary alcohol.
6 0
3 years ago
Choose the reaction that represents the combustion of C6H12O2.
Leona [35]

Answer:

The answer to your question is:  letter D

Explanation:

In a combustion reaction, the reactants are always a molecule with Carbon that reacts with oxygen and the products are carbon dioxide and water.

According to the explanation, the only possible solution is:

a) C₆H₁₂O₂(l) ⇒ 6 C(s) + 6 H₂(g) + O₂(g)

b) Mg(s) + C₆H₁₂O₂(l) ⇒ MgC₆H₁₂O₂(aq)

c) 6 C(s) + 6 H₂(g) + O₂(g) ⇒ C₆H₁₂O₂(l)

d) C₆H₁₂O₂(l) + 8 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(g)

e) None of the above represent the combustion of C₆H₁₂O₂.

4 0
2 years ago
Read 2 more answers
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