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klemol [59]
3 years ago
11

Determine the empirical formula for c6n4o10

Chemistry
1 answer:
ANEK [815]3 years ago
7 0
<h2>Let us derive empirical formula </h2>

Explanation:

We are given with compound C₆N₄O₁₀: mass percentage of all is :

For C = 12 x 6 /288=0.25%

For N= 14 x 4 /288=0.19%

For 0= 16 x 10/288=0.5%

The elements present are :

         Atomic mass   moles                       Simplest ratio           rounding off

C      12                        0.25/12=0.020             0.02/0.02=1           2

N       14                       0.9/14=0.06                    0.06/0.02=3         6

O       16                        0.5/16=0.03                   0.03/0.02=1.5       3

The empirical formula derived is : C₂n₆O₃

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When using 100ml or 50ml graduated cylinder to what decimal place can your volume be estimated?<br>​
Olegator [25]

Answer:

I know that the 100-mL graduated cylinders are always read to 1 decimal place.

I think for 50 mL graduated cylinders, it lets you measure volumes up to 50.0 mL to the nearest 0.1 or 0.2 mL, depending on your exact cylinder.

3 0
3 years ago
What is the mass in grams of 5.00moles of CH4?
anastassius [24]

Answer:

1. 80g

2. 1.188mole

Explanation:

1. We'll begin by obtaining the molar mass of CH4. This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

Number of mole of CH4 from the question = 5 moles

Mass of CH4 =?

Mass = number of mole x molar Mass

Mass of CH4 = 5 x 16

Mass of CH4 = 80g

2. Mass of O2 from the question = 38g

Molar Mass of O2 = 16x2 = 32g/mol

Number of mole O2 =?

Number of mole = Mass /Molar Mass

Number of mole of O2 = 38/32

Number of mole of O2 = 1.188mole

6 0
3 years ago
Why are the oxidation and reduction half-reactions separated in an<br> electrochemical cell?
Aleksandr [31]

Answer:

<h2>It makes the current viable enough to pass through an exterior wire.</h2>

Explanation:

Electrochemical cells primarily comprise of two half-cells. These half-cells assist in isolating the oxidation and reduction half-reactions. These two reactions are linked by a wire which allows the current to move from one edge to the other. The oxidation at the anode and the reduction take place at the cathode and the addition of a salt bridge helps in completing the circuit and permits the current to flow and leads to the generation of electricity.

7 0
2 years ago
I need to find the reactants and products for each of these 7 problems. I need it in 3/5 hours help please!!!
Nezavi [6.7K]

Answer:

The arrow points from the reactants to the products, so just follow the arrows.

Explanation:

some have the reactants on the left and the products on the right, and others are the opposite... just know that

reactants---------> products

or

products<-----------reactants

4 0
3 years ago
What is a major problem with CFCs?
Pepsi [2]
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2 years ago
Read 2 more answers
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