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4 years ago

5

A student prepares 150.0 mL of 1.40 M HCl using 35.0 mL of a stock solution. What is the concentration of the stock solution? Us

e M subscript i V subscript i equals M subscript f V subscript f..

1 answer:

drek231 [11]4 years ago

3 0

Answer:6.0M

Explanation:see attached photo

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15. Why do many substances dissolve easily in water?

mash [69]

Answer:

Water is called the "universal solvent" because it is capable of dissolving more substances than any other liquid. ... Water molecules have a polar arrangement of oxygen and hydrogen atoms—one side (hydrogen) has a positive electrical charge and the other side (oxygen) had a negative charge

Explanation:

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3 years ago

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. How many grams of water would require 4400 joules

Alja [10]

Answer:

The specific heat of water is 4.18 J/g C.

Explanation:

q

=

m

C

s

Δ

T

Never forget that!

2200

=

m

⋅

4.18

J

g

⋅

°

C

⋅

66

°

C

∴

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≈

8.0

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3 years ago

Find the pH of the equivalence point and the volume (mL) of 0.0372 M NaOH needed to reach the equivalence point in the titration

elixir [45]

Answer:

8.54

Explanation:

At equivalence point :

42.2 X 0.052 = Vol. NaOH X 0.0372

Vol of NaOH = 2.1944/0.0372 = 58.99 ml

So volume of NaOH recquired to reach equivalence point = 58.99 ml

Number of miliimoles of CH3COOH = molarity X volume in ml = 42.2 X 0.052 = 2.1944 millimoles

Number of millimoles of NaOH = 58.99 X 0.0372 = 2.1944

Now CH₃COOH and NaOH reacts to give CH₃COONa according to the reaction :

CH₃COOH + NaOH ------> CH₃COONa + H₂O

1 mole of CH₃COOH reacts with 1 mole of NaOH to give 1 mole of CH₃COONa

So 2.1944 millimoles of CH₃COOH will react with 2.1944 millimoles of NaOH to give 2.1944 millimoles of CH₃COONa

So all the acid (CH₃COOH) and base (NaOH) has been converted into salt (CH₃COONa) so there is no acid or base left.

Now molarity of CH₃COONa = number of millimoles of CH₃COONa/total volume in ml = 2.1944/(58.99 + 42.2) = 2.1944/101.19 = 0.02169 M

So using the hydrolysis equation :

pH = 1/2 [ pKw + pKa + log c ]

Ka for acetic acid = 1.75 X 10⁻⁵

so pKa = -log (1.75 X 10⁻⁵) = 4.74

Kw = 10⁻¹⁴

so pKw = -log 10⁻¹⁴ = 14

c = 0.02169

so log c = log 0.02169 = -1.66

putting the values....

pH = 1/2 [14 + 4.74 - 1.66 ]

pH = 1/2 [ 17.08] = 8.54

6 0

3 years ago

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the temperature of a sample of water increases from 20celsius to 46.6celsius as it absorbs 5650 J of heat. what is the mass of t

Levart [38]

Answer:

m = 50.74 kg

Explanation:

We have,

Initial temperature of water is 20 degrees Celsius

Final temperature of water is 46.6 degrees Celsius

Heat absorbed is 5650 J

It is required to find the mass of the sample. The heat absorbed is given by the formula ad follows :

$Q=mc\Delta T$

c is specific heat of water, c = 4.186 J/g°C

So,

$m=\dfrac{Q}{c\Delta T}\\\\m=\dfrac{5650}{4.186\times (46.6-20)}\\\\m=50.74\ kg$

So, the mass of the sample is 50.74 kg.

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TEA [102]

Answer:

the answer is destructive interference

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