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Digiron [165]
3 years ago
12

Earth's original atmosphere probably consisted mostly of which gas?

Chemistry
1 answer:
balandron [24]3 years ago
6 0
D-Carbon Dioxide
Hope this helps :)
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sp2606 [1]

Answer:19. He says that he’s been really tired since several weeks ago. 20. A friend of us is going to pick us up at the airport. 21. I’ve worked like a waiter in the past, but I wouldn’t want to do it again. 22

Explanation:

7 0
2 years ago
Complete combustion of 7.40 g of a hydrocarbon produced 22.4 g of CO2 and 11.5 g of H2O. What is the empirical formula for the h
cluponka [151]
<span>C2H5 First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2. Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087 Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass. moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule. Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen. moles C = 0.50899 moles H = 0.638361 * 2 = 1.276722 We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon. total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185 7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked. Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen. 0.50899 / 1.276722 = 0.398669 0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5. Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is C2H5</span>
3 0
3 years ago
Familiar solutions can have a wide range of pH levels. According to the chart, which is the strongest acid, based on its pH leve
MrRissso [65]

Answer:

Part 1--Gastric acid

Part 2--one hundred times

Part 3--Baking Soda

Explanation:

5 0
3 years ago
Read 2 more answers
Before a bond breaks in a chemical reaction, what happens
Assoli18 [71]
They have to form a chemical bond in order to brake them down first
8 0
3 years ago
400. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution
ch4aika [34]

<u>Answer:</u> The molecular weight of protein is 1.14\times 10^2g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

or,

\pi=i\times \frac{m_{solute}\times 1000}{M_{solute}\times V_{solution}\text{ (in mL)}}}\times RT

where,

\pi = Osmotic pressure of the solution = 0.0861 atm

i = Van't hoff factor = 1 (for non-electrolytes)

m_{solute} = mass of protein = 400 mg = 0.4 g   (Conversion factor:  1 g = 1000 mg)

M_{solute} = molar mass of protein = ?

V_{solution} = Volume of solution = 5.00 mL

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[25+273]K=298K

Putting values in above equation, we get:

0.0861atm=1\times \frac{0.4g\times 1000}{M\times 100}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\M=1136.62g/mol=1.14\times 10^2g/mol

Hence, the molecular weight of protein is 1.14\times 10^2g/mol

4 0
3 years ago
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