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Aneli [31]
3 years ago
5

I really can’t do this don’t understand it

Chemistry
1 answer:
masya89 [10]3 years ago
3 0

Answer:

7.96g, 33.79%

Explanation:

I'll try my best to explain the entire process behind this question ;)

From the question, you can write the reaction

2H_2O(l)->2H_2(g)+O_2(g)

Now, there are a few reasons it is like this. Oxygen is a diatomic element, meaning it doesn't and can't exist as just O. It exists as O₂. To balance, this, double the amount of water and hydrogen so there is an equal amount of  each element on both sides of the reaction (4 H's, 2 O's on the reactant side, and 4 H's, 2 O's on the product side).

From this we can get a mole-to-mole ratio.

Onto the stoichiometry. Our goal in this is to convert from grams of water to grams of hydrogen, and we do so with a mole to mole ratio.

71.0gH_2O*\frac{1molH_2O}{18.02g} *\frac{2molH_2}{2molH_2O}* \frac{2.02g}{1molH_2}\\\\ =7.96gH_2

Basically, what I did was divide by water's molar mass to get moles of water, multiplied by the mole-to-mole ratio (2:2) to get moles of H2, and then multiplied by H2's molar mass to get what should be the amount of H2 produced by the reaction.

For percent yield, you can calculate it is such:

\frac{Actual}{Theoretical}*100

Plug the numbers in:

\frac{2.69g}{7.96g}*100\\\\ =33.79%

So, the percent yield is 33.79%

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Ava’s best friend has an indoor cat. Ava notices that she sneezes every time that she visits her friend. Ava suspects that she i
zlopas [31]

Answer:

I say for her to test her suspicion by going to see an allergist to diagnose the allergy.

7 0
3 years ago
In an experiment, 107.9 grams of H2SO, is produced when 196.2 grams
Ipatiy [6.2K]

Answer:

54.99% yield

Explanation:

percent yield is just the amount you obtained over the amount expected times 100%.

(experimental value/theoretical value) x 100%

= (107.9 g/196.2 g) x 100%

=54.99% yield

5 0
3 years ago
A Downs cell operating at 77.0 A produces 31.0 kg of Na.(a) What volume of Cl₂(g) is produced at 1.0 atm and 540.°C?
nignag [31]

Volume of Cl₂(g) is produced at 1.0 atm and 540.°C=4.5×10^4 L

As per the evenly distributed response

2NaCl (g) ----> 2Na(l)+ Cl2(g)

Calculate the amount of Cl2 that was formed as indicated below:

Moles of Cl2 = 31.0 kg of Na x (1000* 1 * 1 / 1*23* 2)

                   = 673.9 mol

P is equal to 1.0 atm, and T is equal to 813.15 K

when converted to Kelvin by multiplying by a factor of 273.15.

Using Cl2 as an ideal gas, determine the in the following volume:

volume = nRT/P

= 673.9 * 0.0821 * 813.15/ 1

=4.5×10^4 L

As a result, the volume of Cl2 under the given circumstances =4.5×10^4 L

Learn more about Volume here:

brainly.com/question/13338592

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3 0
1 year ago
A sample compound with a molar mass of 34.00g/mol is found to consist of 0.44g H and 6.92g O. Calculate both empirical and molec
balu736 [363]

Answer:

E.F= OH

M.F=O_{2} H_{2}

Explanation:

Empirical Formula

Step 1: Calculate mols of each element

0.44gH(1 mol H/ 1.008g H)= 0.4365 mol H

*note leave extra sig figs for calculations

6.92gO(1 mol O/ 16 g O)=0.4325 mol O

Step 2: Identify which is the smallest mol

*in this case 0.4365 mol H> 0.4325 mol O so we will use 0.4325 mol O

Step 3: Divide above calculations by the smallest mol

0.4325 mol O/0.4325= 1

0.4365 mol H/0.4325= 1.009 *rounds to 1

Step 4: use calculations as subscripts

Oxygen = 1 so the subscript will 1 (O)

Hydrogen = 1 so the subscript will 1 (H)

making E.F= OH

Molecular Formula:

Step 1: identify  molecular mass and mass from the E.F

the molecular mass is given 34.00g/mol

the mass of the E.F is

(oxygen mass from periodic table)+(hydrogen mass from periodic table)

16+1.008= 17.008

Step 2:Divide the molecular mass by the mass given by the emipirical formula.

\frac{34.00}{17.008}= 1.999 round to 2

Step 3:Multiply the empirical formula (the subscripts) by this number to get the molecular formula.  ANSWER: M.F=2(OH)- O_{2} H_{2}

3 0
3 years ago
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