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3 years ago

Why do we say the particles in a rock lying on the ground have kinetic energy and potential energy?

Chemistry

1 answer:

oksano4ka [1.4K]3 years ago

3 0

Answer:

the particles of the rock possess kinetic energy as they stay in a place the particles also contain potential energy due to their position and arrangement This form of stored energy is responsible for keeping the particles together

Explanation:

hop it helps

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Sea water contains roughly 28.0 g of nacl per liter. What is the molarity of sodium chloride in sea water?

Lelechka [254]

Answer:

The molarity of sodium chloride in sea water is 0.479 M

Explanation:

Step 1: Data given

Mass of NaCl = 28.0 grams

Molar mass NaCl = 58.44 g/mol

Step 2: Calculate moles

Moles NaCl = mass NaCl / molar mass NaCl

Moles NaCl = 28.0 grams / 58.44 g/mol

Moles NaCl = 0.479 moles

Step 3: Calculate molarity NaCl in sea water

Molarity = moles / volume

Molarity NaCl = 0.479 moles / 1L

Molarity of NaCl in sea water = 0.479 mol/L = 0.479 M

The molarity of sodium chloride in sea water is 0.479 M

7 0

3 years ago

What is the boiling point of water when the external pressure is 187.5 mmhg?

Lostsunrise [7]

i believe the answer is 65°c

8 0

4 years ago

Which statement identifies the element arsenic?

insens350 [35]

Answer : Option A) Atomic number of Arsenic is 33.

Explanation : Arsenic contains same number of protons in its atomic nucleus. In arsenic there are 33 protons found in the atomic nucleus. Hence, the atomic number will be 33. It has 5 valence electrons in its outermost shell, which is also called as valence shell. So, its valency becomes 5.

8 0

3 years ago

Read 2 more answers

An aluminium kettle weighs 1.05 Kg.

Kipish [7]

Answer:Explained

Explanation:

Given

mass of aluminium kettle( $m_k$ )=1.05 kg

(a)Heat capacity of the kettle=Heat capacity of Aluminium

$C_{Al}=0.9 J/gm-K$

(b)Heat is required to increase the temperature of this kettle from $23^{\circ} to 99 ^{\circ}$

$Q=m_kc_k\Delta T$

$Q=1.05\times 0.9\times 1000\times \left ( 99-23\right )$

Q=71.82 kJ

(c)If kettle contains water 1.25 L of water i.e. 1.246 kg of water then heat required to raise the temperature from 23 to 99

$Q=m_kc_k\left ( \Delta T\right )+m_wc_w\left ( \Delta T\right )$

$Q=1.05\times 0.9\times 1000\times \left ( 99-23\right )+1.246\times 4.184\times 1000\times \left ( 99-23\right )$

Q=71.82+396.20=468.028 kJ

7 0

4 years ago

A student does an experiment to determine the molar solubility of lead(II) bromide. She constructs a voltaic cell at 298 K consi

attashe74 [19]

Answer:

The molar solubility of lead bromide at 298K is 0.010 mol/L.

Explanation:

In order to solve this problem, we need to use the Nernst Equaiton:

$E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}$

E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.

At equilibrium, E = 0, therefore:

$E^{o} = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] = log[ox] - \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] - \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 - \frac{2x5.45x10^{-2} }{0.0591}}\\\\$

[red] = 0.010 M

The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.

7 0

4 years ago