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Cerrena [4.2K]
3 years ago
9

I need help bad Please I need helppoo

Chemistry
1 answer:
Ghella [55]3 years ago
7 0
Its b , with equal force since when he hits the ball . The ball hits back with the same energy
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If A and B are directly proportional and the value of A becomes 3 times as much, what happens to B
dolphi86 [110]
B gets multiplied 3 times as much aswell. (brainly pls) :D
3 0
3 years ago
Read 2 more answers
For an aqueous solution of sodium chloride (NaCl) .Determine the molarity of 3.45L of a solution that contains 145g of sodium
worty [1.4K]
Molar mass NaCl = 58.44 g/mol

number of moles:

mass NaCl / molar mass

145 / 58.44 => 2.4811 moles of NaCl

Volume = 3.45 L

Therefore :

M = moles / volume in liters:

M = 2.4811 / 3.45

M = 0.719 mol/L⁻¹

hope this helps!
6 0
3 years ago
What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i
UNO [17]

Answer:

Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)

The net equation is then:

Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-

Actual cathode half-equation:

Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)

Actual net reaction:

Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)

6 0
3 years ago
You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10 mmol (millimoles) of a
balandron [24]

Answer:

56.9 mmoles of acetate are required in this buffer

Explanation:

To solve this, we can think in the Henderson Hasselbach equation:

pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])

To make the buffer we know:

CH₃COOH  +  H₂O  ⇄   CH₃COO⁻  +  H₃O⁺     Ka

We know that Ka from acetic acid is: 1.8×10⁻⁵

pKa = - log Ka

pKa = 4.74

We replace data:

5.5 = 4.74 + log ([acetate] / 10 mmol)

5.5 - 4.74 = log ([acetate] / 10 mmol)

0.755 = log ([acetate] / 10 mmol)

10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)

5.69 = ([acetate] / 10 mmol)

5.69 . 10 = [acetate] → 56.9 mmoles

6 0
3 years ago
What is the vapor pressure of the pure solvent if the vapor pressure of a solution of 10 g of sucrose (C6H12O6) in 100 g of etha
amid [387]

Answer:

56.4 mmHg

Explanation:

Given:

Vapor pressure of the solution, P solution = 55 mmHg

The mass of sucrose (C₆H₁₂O₆) = 10 g

Also, Molar mass of sucrose (C₆H₁₂O₆) = 180 g/mol

So, moles = Given mass/ molar mass

Hence, moles of sucrose in the solution = 10 g / 180 g/mol = 0.05556 mol

Given that: Mass of ethanol = 100 g

Molar mass of ethanol = 46 g/mol

Hence, moles of ethanol = 100 g / 46 g/mol = 2.174 mol

Mole fraction of solvent, ethanol is:

X ethanol = 2.174 mol / (2.174 + 0.05556) mol = 0.975

Applying Raoult's Law

P solution = X ethanol*P° ethanol

<u> => P° ethanol  = P solution / X ethanol  = 55 mmHg / 0.975 = 56.4 mm Hg</u>

6 0
4 years ago
Read 2 more answers
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