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Advocard [28]
4 years ago
10

Rhhhhhh hhdjasdlk lksdjdklsj askdj ​

Chemistry
2 answers:
DochEvi [55]4 years ago
8 0

how is this a question

statuscvo [17]4 years ago
5 0

Answer:

hehsn3eyn8yehuy3bdyrd

juusuususuusuusu

Explanation:

You might be interested in
An electrochemical cell is constructed using electrodes with the following half-cell reactions.
marshall27 [118]

Answer:

E cell = +1.95 V

Explanation:

At Anode : Oxidation reaction takes place

At Cathode : Reduction reaction takes place

The reaction with lower value of reduction potential will undergo Oxidation

Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s)   E = -1.18 V

This equation undergo oxidation reaction and become:

Anode(Oxidation-Half) :

Mn(s)\rightarrow Mn^{2+}(aq)+2e^{-}      E = +1.18 V

Cathode(Reduction-Half) :

Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}    E =+0.77 V

To balance the reaction multiply reduction-Half with 2.We get :

Fe^{3+} +2Mn(s)+\rightarrow Fe^{2+} + 2Mn^{2+}

Note that E is intensive property , do not multiply E  of oxidation-half with 2

Ecell =  0.77 -(-1.18)

E = +1.95 V

3 0
3 years ago
At a constant pressure, a sample of gas occupies 4.2 L at 210 K. What volume will the gas occupy at 260 K?*
yan [13]

Answer:

V1 / T1 = V2 / T2

V₁ = 420 mL

T₁ = 210 K

V₂ = ?

T₂ = 250 K

420 / 210 = V2 / 250

V2 = (420 / 210) x 250 = 500 mL

Explanation:

6 0
3 years ago
What is the empirical formula of a compound which contains 84.4% c and 15.6% h by mass?
podryga [215]

Answer: The empirical formula for the given compound is CH_2

Explanation : Given,

Percentage of C = 84.4 %

Percentage of H = 15.6 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 84.4 g

Mass of H = 15.6 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{84.4g}{12g/mole}=7.03moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{15.6g}{1g/mole}=15.6moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.03 moles.

For Carbon = \frac{7.03}{7.03}=1

For Hydrogen  = \frac{15.6}{7.03}=2.22\approx 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 2

Hence, the empirical formula for the given compound is C_1H_2=CH_2

7 0
3 years ago
Using the chemical formula CaO + 2 NaCl → Na2O + CaCl2, calculate how many moles of CaCl2 you would produce if you used up 5.4 m
murzikaleks [220]

Answer:

5.4 moles CaO   1 mole CaCl₂   5.4 mole CaCl₂

                           1 mole CaO

Explanation:

7 0
2 years ago
A 57 g g sample of iron reacts with 24 g g of oxygen to form how many grams of iron oxide?
MatroZZZ [7]
Your answer will be 80g
3 0
3 years ago
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