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2 years ago

What is [h3o ] in a solution of 0.25 m ch3co2h and 0.030 m nach3co2?

1 answer:

4 0

Hello!

To determine [H₃O⁺], we need to apply the Henderson-Hasselback equation, since this is a case of an acid and its conjugate base:

$pH=pKa+log( \frac{[A^{-}] }{[HA]} )$

$pH=4,76+log( \frac{0,030M}{0,25M} ) \\ \\ pH=3,84$

Now, we use the definition of pH and clear [H₃O⁺] from there:

$pH=-log[H_3O^{+}]$

$[H_3O^{+}] = 10^{-pH} =10^{-3,84}=0,00014 M$

So, the [H₃O⁺] concentration is 0,00014 M

Have a nice day!

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A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro

Savatey [412]

Answer:

The correct answer is option C.

Explanation:

1.0 g sample of a cashew :

Heat released on combustion of 1.0 gram of cashew = -Q

We have mass of water = m = 1000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q

$Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J$

Heat released on combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows :

Heat released on combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q'

$Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J$

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

$q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J$

Heat released on combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

5 0

2 years ago

How many moles of N2 are essential for generating 0.08 moles of Li3N in the given reaction?6Li N2 2Li3N

jeka57 [31]

08 moles Li3N * 1mole N2/2moles Li3N = 0.04

3 0

2 years ago

Read 2 more answers

A mixture of gases A2 and B2 are introduced to a slender metal cylinder that has one end closed and the other fitted with a pist

REY [17]

Answer:

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

Explanation:

Determination of sign of q

Temperature of the water bath before the reaction = 25 °C

Temperature of the water bath after the completion of the reaction = 28 °C

After the completion of the reaction, temperature of the water bath is increased that means heat is released during the reaction and flows out of the system.

If heat is absorbed by the system, then q is indicated by positive sign and if heat is released by the system, then q is indicated by negative sign.

As in the given case, heat is released by the system, so sign of q is negative, or q < 0

Determination of sign of w

After the completion of the reaction, piston moved downward, that means volume of the system decreases or compression occur. During the compression, work is done on the system.

if work is done on the system, sign of w is positive.

If work is done by the system, sign of w is negative.

In the given case, work is done on the system, therefore sign of w is positive, or w > 0

Determination of sign of ΔE

Relationship between ΔE, q and w is given by first law of thermodynamics:

ΔE = q + w

In this case, q is positive and w is negative, so the sign of ΔE depends of magnitude of q and w. As magnitude of w and q cannot be determined in this case, thus, the given information is insufficient for the determination of sign of ΔE.

So, among the given option, option c is correct.

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

3 0

2 years ago

In the third period of the periodic table sodium is followed by magnesium aluminum silicon and phosphorus which of these element

wariber [46]

Answer:
Phosphorous has the smallest atomic size.

Explanation:
As we know these elements belong to same period means there valence shell is the same. So moving from left to right along the period the shell number remains constant but the number of protons and electrons increases. So, due to increase in number of protons the nuclear charge increases hence attracts the valence electrons more effectively resulting in the decrease of atomic size.

Elements and their atomic radius are as follow,

Magnesium 0.160 nm

Aluminium 0.130 nm

Silicon 0.118 nm

Phosphorus 0.110 nm

6 0

2 years ago

During this reaction, p4 + 5o2 → p4o10, 0.800 moles of product was made in 15.0 seconds. what is the rate of reaction? 56.8 g/mi

Eduardwww [97]

The solution for this problem would be:
The mass of P4O10 is computed by: 0.800 mol x 284 g/mol = 227g t = 15.0 s ( 1 min / 60 s) = 0.25 min
So solving for the rate will be mass over t = m/t = 227/0.25 = 908 g/min would be the answer for this problem.

3 0

2 years ago