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yuradex [85]
3 years ago
6

1. A student mixes two liquids together and notices that a white solid has formed

Chemistry
1 answer:
monitta3 years ago
7 0

Answer:

3. A chemical reaction has occured

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What is the molarity of a solution made by dissolving 14.8 g of ammonium hydroxide NH4OH, in enough water to make 250.0 mL of so
kicyunya [14]

Answer:

Molarity= 1.69M

Explanation:

m= 14.8, Mm= 35, V= 0.25dm3, C= ?

Moles = m/M= C×V

Substitute and Simplify

m/M= C×V

14.8/35= C×0.25

C= 1.69M

6 0
2 years ago
How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? 4 KNO3(s) → 2 K2O(s) + 2
Leya [2.2K]

Answer:

0.725 mol

Explanation:

Moles are calculated as the given mass divided by the molecular mass.

i.e. ,

moles = ( mass / molecular mass )

since,

mass of KNO₃ = 58.6 g  ( given )

Molecular mass of KNO₃ = 101 g / mol

Therefore,

moles of KNO₃ = 58.6 g / 101 g / mol

moles of KNO₃ = 0.58 mol

From the balanced reaction ,

4 KNO₃ (s) ---> 2K₂O (s) + 2N₂ (g) + 5O₂ (g)

By the decomposition of 4 mol of KNO₃ , 5 mol of O₂ are formed ,

hence, unitary method is used as,

1  mol of KNO₃  gives 5 / 4 mol O₂

Therefore,

0.58 mol of KNO₃ , gives , 5 / 4  * 0.58 mol of O₂

Solving,

0.58 mol of KNO₃ , gives , 0.725 mol of O₂

Therefore,

58.6g of KNO₃ gives 0.725 mol of O₂.

3 0
2 years ago
For the Earth's atmosphere, section number two in the pie graph BEST represents the percentage of A) argon B) oxygen C) nitrogen
larisa86 [58]

-<u><em>Oxygen</em></u>

According to Google these are the percentages of the <em>Earths Atmosphere</em>

<em>1</em> 78% - Nitrogen

<u>2</u> 21% - Oxygen

<em>3</em> 0.9% - Argon

<em>4 </em>0.3 - Carbon Dioxide with very small percentage of other elements.

3 0
3 years ago
Using the Lewis dot structures of magnesium and oxygen, predict the ionic formula
Blizzard [7]

Answer:

(edit: nvm I figured it out, here is the answer)

Explanation:

7 0
3 years ago
A concentration cell consisting of two hydrogen electrodes (PH2 = 1 atm), where the cathode is a standard hydrogen electrode and
Lunna [17]

The pH of the unknown solution is 3.07.

<u>Explanation:</u>

<u>1.Find the cell potential as a function of pH</u>

From the Nernst Equation:

Ecell=E∘cell−RT /zF × lnQ

where

R denotes the Universal Gas Constant

T denotes the temperature

z denotes the moles of electrons transferred per mole of hydrogen

F denotes the Faraday constant

Q denotes the reaction quotient

Substitute the values,

E∘cell=0   lnQ=2.303logQ

E0cell=−2.30/RT /zF × log Q

Solving the equation,

<u>2. Find the Q  value</u>

Q=[H+]2prod pH₂, product/ [H+]2reactpH₂, reactant

Q=[H+]^2×1/1×1=[H+]2

Taking the log

logQ= log[H+]^2=2log[H+]=-2pH

From the formula,

Ecell=−2.303RT /zF× logQ

E cell= 2.303 × 8.314 CK mol (inverse)  × 298.15

K × 2pH /2×96 485  C⋅mol

( inverse)

E cell= 0.0592 V × pH

<u>3. Finding the pH value</u>

E cell= 0.0592 V × pH

pH = E cell/ 0.0592 V= 0.182V/ 0.0592V

pH=3.07

The pH of the unknown solution is 3.07.

7 0
3 years ago
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