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denis23 [38]
2 years ago
14

When the fish water had pH of 8.0, the hydronium ion concentration is 1.0x 10^-8 mole per liter. What is Hydornium ion con ent r

ation when water had pH of 7.0
Chemistry
1 answer:
Klio2033 [76]2 years ago
4 0

The hydronium ion concentration when pH = 7 is :  8.75 * 10⁻⁹ mole/liter

<h3>Determine hydronium ion concentration when pH = 7 </h3>

Given that :

pH = 8.0

Hydronium ion concentration = 1 * 10⁻⁸ mole/liter

Resolving the question

8.0 = 1 * 10⁻⁸ mole/liter

7.0 =  x

therefore :

x = 7 ( 1 * 10⁻⁸ ) / 8.0

  = 8.75 * 10⁻⁹ mole/liter

Hence we can conclude that The hydronium ion concentration when pH = 7 is :  8.75 * 10⁻⁹ mole/liter

Learn more about hydronium ion : brainly.com/question/27038951

#SPJ1

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Calculate the number of molecules in 0.75 moles of CH₄. x 10²³ molecules
spin [16.1K]

<u>Ans :</u>4.5 * 10²³ molecules

<u>Given:</u>

Moles of CH4 = 0.75

<u>To determine;</u>

The number of molecules of CH4 in 0.75 moles

<u>Explanation:</u>

1 mole of CH4 contains Avogadro's number i.e. 6.023*10²³ molecules

Therefore, 0.75 moles would correspond to :

= 0.75 moles * 6.023*10²³ molecules/1 mole

= 4.5 * 10²³ molecules

7 0
2 years ago
A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad
ASHA 777 [7]

0 \; \textdegree{\text{C}}

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

  • E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ} and
  • m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}

Let the final temperature of the system be t \; \textdegree{\text{C}}. Thus \Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial})  - t_{0} = t \; \textdegree{\text{C}}

  • Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ} (converted to kilojoules)
  • Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}
  • Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable t.

E(\text{absorbed} ) = E(\text{released})

  • E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})
  • E(\text{released}) =  Q(\text{lead})

Confirm the uniformity of units, equate the two expressions and solve for t:

66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)

t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}} which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., t \le 0 \; \textdegree{\text{C}}. However, there's no way for the temperature of the system to go below 0 \; \textdegree{\text{C}}; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at 0 \; \textdegree{\text{C}}; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

3 0
3 years ago
Please help due today
Murljashka [212]
The answer for the question is c.
7 0
2 years ago
In the laboratory a student finds that it takes 44.0 Joules to increase the temperature of 10.6 grams of solid zinc from 24.9 to
BaLLatris [955]

Answer:

The specific heat of zinc is 0.361 J/g°C

Explanation:

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

44.0 J needed

Mass of solid zinc = 10.6 grams

Initial temperature = 24.9 °C

Final temperature = 36.4 °C

<u>Step 2</u>: Calculate the specific heat of zinc

Q = m*c*ΔT

⇒ with Q = heat (in Joule) = 44.0 J

⇒ with m = the mass of the solid zinc = 10.6 grams

⇒ with c = the specific heat of the zinc = TO BE DETERMINED

⇒ with ΔT = The change in temperature = T2-T1 = 36.4 °C - 24.9 °C = 11.5 °C

44.0 J = 10.6 grams * c * 11.5°C

c = 44.0 J / (10.6g * 11.5 °C)

c = 0.361 J/g°C

The specific heat of zinc is 0.361 J/g°C

7 0
3 years ago
Which of the following atoms has 4 valence electrons
Mars2501 [29]
Can you show me the atoms please? I would be able to help.
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