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3 years ago

What is the percent by mass of carbon in acetone, c 3 h 6 o?

Chemistry

2 answers:

Angelina_Jolie [31]3 years ago

6 0

Answer: I just took the test and got 100%

1. B: 0.430 mol

2. B: percent composition

3. C: 71.5%

4. B: 62.1%

5. A: empirical formula

6. D: C16H24O4

7. D: 673g

Feel free to mark as brainliest :)

Tatiana [17]3 years ago

5 0

In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).

Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol

To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C= $( \frac{3(12 g/mol)}{3(12 g/mol)+6(1 g/mol)+16 g/mol} ) x 100%$
%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%

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2 years ago

What are the roles of producers, primary consumers, secondary consumers, tertiary consumers for the trophic levels in the energy

Nesterboy [21]

This question is more for Biology than Chemistry, but the role of producers is to make energy (food) to be consumed. In a pyramid diagram, the producers would be at the bottom. Now going up the pyramid, the primary conumers are the first to consume producers and obtain energy from them. As you go up the pyramid, the secondary consumers will consume the primary consumers as a way to obtain energy, and the same goes for tertiary consumers towards secondaries.

As you go up the energy pyramid, you will notice a trend that there is less energy being obtained from each consumer. In other words, the producers will ALWAYS have more energy than the tertiary consumers.

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3 years ago

Calculate δ h for the reaction:no (g) + o2 (g) ↔ no2 (g). given: 2o3(g) ↔ 3o2(g) δh=-426 kj o2(g) ↔ 2o(g) δh=+ 490 kj no(g) + o3

maks197457 [2]

To calculate the δ h, we must balance first the reaction:

NO + 0.5O2 -----> NO2

Then we write all the reactions,

2O3 -----> 3O2 δ h = -426 kj eq. (1)

O2 -----> 2O δ h = 490 kj eq. (2)

NO + O3 -----> NO2 + O2 δ h = -200 kj eq. (3)

We divide eq. (1) by 2, we get

O3 -----> 1.5O2 δ h = -213 kj eq. (4)

Then, we subtract eq. (3) by eq. (4)

NO + O3 -----> NO2 + O2 δ h = -200 kj
- (O3 -----> 1.5 O2 δ h = -213 kj)
NO -----> NO2 - 0.5O2 δ h = 13 kj eq. (5)

eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)

O -----> 0.5O2 δ h = -245 kj eq. (6)

Add eq. (6) to eq. (5), we get

NO -----> NO2 - 0.5O2 δ h = 13 kj
+ O -----> 0.5O2 δ h = -245 kj
NO + O ----> NO2 δ h = -232 kj

ANSWER: NO + O ----> NO2 δ h = -232 kj

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3 years ago

: An unknown metal crystallizes in the cubic crystal structure. The metal has a radius 140pm, atomic mass of 135 g/mol, and dens

ki77a [65]

The cubic unit cell this metal crystallize as is BCC structure .

The structure of a crystalline solid, whether a metal or not, is best described by considering its simplest repeating unit, which is referred to as its unit cell.

The unit cell consists of lattice points that represent the locations of atoms or ions.

The entire structure then consists of this unit cell repeating in three dimensions

$\rm \rho =\dfrac{nM}{N_{0} a^{3}} \\\\\\\\\rm n =\dfrac{\rho N_{0} a^{3}}{M} \\\\\\\\Assuming\; it \; to \; be \; a\; BCC\; structure \\\\\\ r = \dfrac{\sqrt{3} \times a}{4} \\\\\\Therefore\; a = 3.23 \times 10^{-8}\\\\\\\\\rm n =\dfrac{13.3 \times 6.022 \times 10^{23}\times 3.23^{3}\times (10^{-8})^{3}}{135} \\\\$

n= 2

Hence our assumption was correct

It is a BCC structure .

Therefore the cubic unit cell this metal crystallize as is BCC structure .

To know more about unit cell

brainly.com/question/13110055

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The collision of the molecules between the hydrogen molecule or H2, and an iodine molecule or I2, provided there would be a sufficient energy is that the system would eventually undergo a chemical change wherein a new chemical compound would be formed from these two molecules.

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