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Kisachek [45]
3 years ago
10

7. What measures are applicable in the context of Nepal to push energy crisis further. Describe any three points clearly.​

Chemistry
1 answer:
Tamiku [17]3 years ago
4 0

Answer:

which energy?........

Explanation:

..........................

........

....

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A beaker with 1.60×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and c
stich3 [128]

Answer:

The pH will change 0.16 ( from 5.00 to 4.84)

Explanation:

Step 1: Data given

volume of acetic acid buffer = 160 mL

The total molarity of acid and conjugate base in this buffer is 0.100 M

A student adds 7.10 mL of a 0.460 M HCl solution to the beaker.

The pKa of acetic acid is 4.740

pH = 5.00

Step 2: Calculate concentration of acid

Consider x = concentration acid

Consider y = concentration conjugate base

x + y = 0.100

5.00 = 4.740 + log y/x

5.00 - 4.740 = log y/x

0.26 = log y/x

10^0.26 =1.82 = y/x

1.82 x = y

Since x+y = 0.100

x + 1.82 x = 0.100

2.82 x = 0.100

x =0.0355 M = concentration acid

Step 3: Calculate concentration of conjugate base

y = 0.100 - x

0.100 - 0.0355 =0.0645 M= concentration conjugate base

Step 4: Calculate moles of acid

Moles = volume * molarity

moles acid = 0.160 L * 0.0355 M= 0.00568  moles

Step 5: Calculate moles of conjugate base

moles conjugate base = 0.0645 M * 0.160 L=0.01032 moles

Step 6: Calculate moles HCl

moles HCl = 7.10 * 10^-3 L * 0.460 M=0.003266 moles

Step 7: Calculate new moles

A- + H+ = HA

moles conjugate base = 0.01032 - 0.003266 =0.007054  moles

moles acid = 0.00568 + 0.003266=0.008946 moles

Step 8: Calculate the total volume

total volume = 160 + 7.10 = 167.1 mL = 0.1671 L

Step 9: Calculate the concentration of the acid

concentration acid = 0.008946/ 0.1671 =0.0535 M

Step 10: Calculate the concentration of conjugate base

concentration conjugate base = 0.007054/ 0.1671 =0.0422 M

Step 11: Calculate the pH

pH = 4.740 + log 0.0535/ 0.0422=4.84

change pH = 5.00 - 4.84=0.16

The pH will change 0.16

5 0
3 years ago
How many milliliters of a 0.1000 M NaOH solution would be needed to neutralize a 1.0000-g sample of potassium hydrogen phthalate
7nadin3 [17]

Answer: 2) 2HCl(sq) + CaCO3(s) CaCl2(sq) + CO2(g) + H2O (l) No of moles of CaCO3 = amount of the CaCO3 (g)/mw of CaCO3 (g/mole)= 0.8085 g/100 g/mole = 0.008085

Explanation:

7 0
2 years ago
Name the acids present in Tamarind Tomato and Vinegar?​
OverLord2011 [107]

Answer:

<h3>tartaric acids</h3>

The molecular formula of the citric acid is C6H8O7. The structure of citric acid is as follows: The acid present in tamarind is tartaric acid.

<h3>Vinegar- </h3>

Oxalic

3 0
3 years ago
Read 2 more answers
When the pressure and number of particles are kept constant for a sample of gas, which of the following is also constant for the
Leno4ka [110]
The answer is D....................
7 0
3 years ago
How many grams of Co are needed to react with an excess of fe203 to produce 156.2 g FE? show your work.
Gekata [30.6K]

The reaction is given as

Fe2O3 (s)+ 3CO(g)--->3CO2(g)+ 2Fe(s)

No.of moles=mass in gram/molar mass

As for Fe mole =156.2g/55.847=2.7969~2.797

The ratio b/w CO and Fe is 3:2

Moles of CO needed= 2.797x3/2=4.1955

Mass of CO needed= 4.195mol x 28.01g/mol= 117.515g

8 0
3 years ago
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