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Art [367]
4 years ago
14

Why isn't the Liter a basic SI unit of measurement? (In your answer use at least one SI base unit of measurement and one other d

erived unit of measurement to support your answer.) *
Chemistry
1 answer:
Y_Kistochka [10]4 years ago
5 0

Answer:

What give me a few minutes I have to finish my test them I will answer in comments.

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How do volcanoes cause changes in the Earth's surface?
scoundrel [369]

Answer: If lava hit water it forms rock that helps make islands and moutains

Explanation:

5 0
3 years ago
Read 2 more answers
What is the percent mass oxygen in calcium carbonate (CaCo3)?
Gnoma [55]
  <span>Step 1 is to determine the mass of each part 
Mass of Ca is 40.08 g 
Mass of C is 12.01 g 
Mass of O is 16.00 x 3 = 48.00 g 
Step 2 is to determine the total mass of the compound 
Total mass of CaCO3 is 40.08 + 12.01 + 48.00 = 100.09 g 

Step 3 is to determine the % of each part using the following formula: 
Mass of part / total mass x 100 = 

40.08 / 100.09 x 100 = 40.04 % Ca 

12.01 / 100.09 x 100 = 12.00 % C 

48.00 / 100.09 x 100 = 47.96 % O 

Step 4 is to double check by adding all percentages. If they equal 100, then I probably did it right. :) 
40.04 
+12.00 
+47.96 
=100.00</span><span>
</span>
4 0
3 years ago
Read 2 more answers
The element X has three naturally occurring isotopes. The masses (amu) and % abundances of the isotopes are given in the table b
denis-greek [22]

Answer:

220.42098 amu

Explanation:

(220 .9 X  .7422) + (220 X .0.1278) + (218.1 X 0.13) =     220.42098 amu

These are weighted averages.

So, we will take mass of one and multiply by abundance percentage that is provided and add them together.

In order to calculate the  average atomic mass, we have to convert the percentages of abundance to decimals. So, you get

(220 .9 X  .7422) + (220 X .0.1278) + (218.1 X 0.13) =     220.42098 amu

6 0
3 years ago
How does a graph with no trend differ from a graph with anomalous data points?
Gelneren [198K]
It differs because it shows different volume of information and it gives a different point of indicanton of numbers.
4 0
3 years ago
Read 2 more answers
Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi
Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g)     ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

[]=\frac{P}{R.T}

<em>Calculate ΔG at 25°C given the following sets of partial pressures.</em>

<em>Part A  130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.</em>

[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M

[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

<em>Part B  5.0atm SO₂, 3.0atm O₂, 30atm SO₃  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M

[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M

[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

<em>Part C Each reactant and product at a partial pressure of 1.0 atm.  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0
3 years ago
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