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pickupchik [31]
3 years ago
11

A different solution with an H+ concentration of 1.0 × 10–4 would have a pH = .

Chemistry
1 answer:
tekilochka [14]3 years ago
7 0
1×10^-4 = 0,0001M

pH = -log[H+]
pH = -log0,0001
pH = 4
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potassium, k, has an atomic number of 19 and an atomic mass of 39 how many electrons does a neutral atom of K have?
coldgirl [10]
A neutral atom of potassium has 19 electrons.
8 0
3 years ago
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kotegsom [21]
Im pretty sure it would be d.
7 0
3 years ago
Read 2 more answers
A 1.8 g sample of octane C8H18 was burned in a bomb calorimeter and the temperature of 100 g of water increased from 21.36 C to
melomori [17]

Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

Explanation:

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

In other words, 3101.56 J of heat was evolved from the reaction of 1.8 g octane with water.

Heat of combustion of octane per gram:

1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

= 1723.08 J of heat

So, heat of combustion of octane per gram is 1723.08 J

Heat of combustion per mole:

1.8 g of octane produces 3101.56 J of heat

1 mole of octane will produce X J of heat

1 mole of octane = 114 g/ mol of octane

So we have:

1.8 g of octane = 3101.56 J

114 g of octane = (3101.56 * 114 / 1.8) J of heat

= 196 432.13 J

= 196. 4 kJ of heat

The heat of combustion of octane per mole is 196.4 kJ /mol.

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

8 0
3 years ago
Maressa has 150 g of a NaCl solution that has concentration of 15.0 % NaCl (mm). How much NaCl is dissolved in that solution? O
Iteru [2.4K]

Answer:

I don't know if option e. is 22,5 g. because that is the correct answer for this. 15% NaCl (mm) gives you the information that in 100 g of solution, you have 15 g of solute. So, if in 100 g of solution you have 15 g of NaCl, in 150 g of solution, how much. Try the rule of three.

Explanation:

4 0
3 years ago
Consider 4.00 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.80 L and the temperature is increased to 3
stealth61 [152]

This is an exercise in<u> the General Combined Gas Law</u>.

To start solving this exercise, we obtain the following data:

<h3>Data:</h3>
  • V₁ = 4.00 l
  • P₁ = 365 mmHg
  • T₁ = 20 °C + 273 = 293 K
  • V₂ = 2,80 l
  • T₂ = 30 °C + 273 = 303 K
  • P₂ = ¿?

We apply the following formula:

  • P₁V₁T₂=P₂V₂T₁     ⇒  General formula

Where:

  • P₁=Initial pressure
  • V₁=Initial volume
  • T₂=end temperature
  • P₂=end pressure
  • T₂=end temperature
  • V₁=Initial temperature

We clear for final pressure (P2)

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{P_{1}V_{1}T_{2}}{V_{2}T_{1}} \ \ \to \ \ \ Formula \end{gathered}$}

We substitute our data into the formula:

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{(365 \ mmHg)(4.00 \not{l})(303 \not{K})}{(2.80 \not{l})(293\not{K})}  \end{gathered}$}

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{442380 \ mmHg}{ 820.4 }  \end{gathered}$}

\boxed{\large\displaystyle\text{$\begin{gathered}\sf P_{2}=539.224 \ mmHg \end{gathered}$}}

Answer: The new canister pressure is 539.224 mmHg.

<h2>{ Pisces04 }</h2>
6 0
2 years ago
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