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2 years ago

Which conversion factor is needed to solve the following problem?

Chemistry

1 answer:

USPshnik [31]2 years ago

3 0

Molar Volume is required to solve this problem. As we know that "1 mole of any gas at standard temperature and pressure occupies 22.4 L of volume". SO using this concept, we can calculate the volume of ammonia formed by reacting 54.1 L of Hydrogen gas as follow,

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Which of these has a higher hydrogen ion concentration? *

Genrish500 [490]

Answer:

pOH = 8.9

Explanation:

pOH+pH = 14

pH= 14- pOH

pOH = 8.9, pH=14-8.9 = 5.1

pOH = 5.5, pH=14-5.5 = 8.5

Than less pH , than more H+ ion concentration.

pH=5.1 <pH=8.5

pH=5.1 is more acidic solution

pH=5.1 or pOH=8.9 has a higher H+ concentration.

8 0

2 years ago

6) Andy has left a bucket of smelly cheese porridge to cool down in the

natta225 [31]

Explanation:

the air particales are always moving so it also takes the smell of the chesse with it and sreads everywhere in the house.

7 0

2 years ago

Which subatomic particles would be found orbiting around the nucleus?

LiRa [457]

An atom consists of three subatomic particles.

In the middle of atom it consists of nucleus which have two subatomic particles. They are protons and neutrons.

And there is a subatomic particle that is found orbiting around the nucleus in an atom, name of that subatomic particle is electron.

Hope this helps!

5 0

2 years ago

A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

nikitadnepr [17]

Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$

3 0

2 years ago

What mass of potassium chloride, KCl, is produced when 12.6 g of oxygen, 02, is produced?

blagie [28]

Mass of KCl= 19.57 g

<h3>Further explanation</h3>

Given

12.6 g of Oxygen

Required

mass of KCl

Solution

Reaction

2KClO3 ⇒ 2KCl + 3O2

mol O2 :

= mass : MW

= 12.6 : 32 g/mol

= 0.39375

From the equation, mol KCl :

= 2/3 x mol O2

= 2/3 x 0.39375

=0.2625

Mass KCl :

= mol x MW

= 0.2625 x 74,5513 g/mol

= 19.57 g

4 0

1 year ago