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irina1246 [14]
3 years ago
10

if i add water to 100 mL of a 0.75 M NaOH solution un til the final volume is 165mL what will the molarity of the diluted soluti

on be
Chemistry
1 answer:
ehidna [41]3 years ago
7 0
Answer is: <span>the molarity of the diluted solution 0,454 M.
</span>V₁(NaOH) = 100 mL ÷ 1000 mL/L = 0,1 L.
c₁(NaOH) = 0,75 M = 0,75 mol/L.
n₁(NaOH) = c₁(NaOH) · V₁(NaOH).
n₁(NaOH) = 0,75 mol/L · 0,1 L.
n₁(NaOH) = 0,075 mol
n₂(NaOH) = n₁(NaOH) = 0,075 mol.
V₂(NaOH) = 165 mL ÷ 1000 mL/L = 0,165 L.
c₂(NaOH) = n₂(NaOH) ÷ V₂(NaOH).
c₂(NaOH) = 0,075 mol ÷ 0,165 L.
c₂(NaOH) = 0,454 mol/L.
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Hii can someone help me pls this is the last question
lara31 [8.8K]

Answer:

B. 126

Explanation:

the scale is 1 to 42 so the scale would be:

3 cm:x cm

3 * 42 = 126 cm

8 0
2 years ago
Using a 300 MHz NMR instrument:
Softa [21]

Answer:

a. 750Hz, b. 4.0ppm, c. 600Hz

Explanation:

The Downfield Shift (Hz) is given by the formula

Downfield Shift (Hz) = Chemical Shift (ppm) x Spectrometer Frequency (Hz)

Using the above formula we can solve all three parts easily

a. fspec = 300 MHz, Chem. Shift = 2.5ppm, 1MHz = 10⁶ Hz, 1ppm (parts per million) = 10⁻⁶

Downfield Shift (Hz) = 2.5ppm x 300MHz x (1Hz/10⁶MHz) x (10⁻⁶/1ppm)

Downfield Shift = 750 Hz

The signal is at 750Hz Downfield from TMS

b. Downfield Shift = 1200 Hz, Chemical Shift = ?

Chemical Shift = Downfield shift/Spectrometer Frequency

Chemical Shift = (1200Hz/300MHz) x (1ppm/10⁻⁶) = 4.0 ppm

The signal comes at 4.0 ppm

c. Separation of 2ppm, Downfield Shift = ?

Downfield Shift (Hz) = 2(ppm) x 300 (MHz) x (1Hz/10⁶MHz) x (10⁻⁶/1ppm) = 600 Hz

The two peaks are separated by 600Hz

6 0
3 years ago
A 100.00 ml of volume of 0.500 M HCl was mixed with 100.00 ml of 0.500 M KOH in a constant pressure calorimeter. The initial tem
Virty [35]

Answer:

The heat of the reaction = -1985 J = -1.985 kJ

The enthalpy of the reaction is -39.7 kJ/ mol

Explanation:

<u>Step 1: </u>Data given

Volume of HCl = 100 mL the heat of the reaction = 0.1 L

Molarity of HCl solution = 0.500 M

Volume of KOH = 100 mL = 0.1 L

Molarity of KOH solution = 0.500 M

Initial temperature = 23.0 °C

Final temperature = 25.5 °C

Specific heat of the solution = 3.97 J/°C *g

Density of the solution = 1g/ mL

<u>Step 2: </u>Calculate heat

Q = m*c*ΔT

with m = the mass of both solution : 100g + 100 g ( since density = 1g/mL) = 200 g

c = the specific heat = 3.97 J/°C*g

ΔT  = T2 -T1 = 25.5 = 23 = 2.5 °C

Q = 200g *3.97 J/°C*g * 2.5°C = 1985 J  (= -1985 J because it's exothermic)

<u />

<u>Step 3:</u> Calculate the number of moles

Number of moles = Molarity * Volume

Number of moles = 0.5 * 0.1 L = 0.05 moles

(Moles of the acid are equal to the moles of water produced.

Moles of solution = 0.05 moles)

<u>Step 4: </u>Calculate the enthalpy of the reaction

ΔH = heat change /Number of moles

    = -1.985 kJ/ 0.05 moles

   =- 39700 J/mol = -39.7 kJ/ mol

The enthalpy of the reaction is -39.7 kJ/ mol

The enthalpy of the reaction is negative, this means the reaction is exothermic ( which means the final temperature is higher than the initial temperature.)

7 0
3 years ago
A steel tank contains carbon dioxide at a pressure of 13.0 atm when the temperature is 34oC. What will be the internal gas press
morpeh [17]

Answer:

D. 15.8atm

Explanation:

Given parameters:

Initial pressure = 13atm

Initial temperature  = 34°C = 34 + 273  = 307K

Final temperature  = 100°C = 100 + 273  = 373K

Unknown:

Final pressure  = ?

Solution:

To solve this problem, we apply a derivation of the combined gas law taking the volume as a constant.

The expression is shown mathematically below;

        \frac{P_{1} }{T_{1} }   = \frac{P_{2} }{T_{2} }

P and T pressure and temperature values

1 and 2 are initial and final states

 Insert the parameters and solve for T₂;

    \frac{13}{307}   = \frac{P_{2} }{373}  

        P₂  = 15.8atm

4 0
3 years ago
In the absence of sodium methoxide, the same alkyl bromide gives a different product. Draw an arrowpushing mechanism to account
hoa [83]

Answer:

See explanation below

Explanation:

The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.

Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.

For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)

For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.

3 0
3 years ago
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