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tatuchka [14]
3 years ago
14

How many moles are in 1.50 grams of ethanol, CH3CH2OH?

Chemistry
2 answers:
LekaFEV [45]3 years ago
4 0
CH3CH2OH = 46.068 (molar mass)
1.50g(1 mole / 46.0680) =
0.03256 moles
Galina-37 [17]3 years ago
4 0

<u>Answer:</u> The number of moles of ethanol for given amount is 0.0326 moles.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We are given:

Given mass of ethanol = 1.50 g

Molar mass of ethanol = 46.07 g/mol

Putting values in above equation, we get:

\text{Moles of ethanol}=\frac{1.50g}{46.07g/mol}=0.0326mol

Hence, the number of moles of ethanol for given amount is 0.0326 moles.

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Li2SO4 _____ an electrolyte in solution.<br> A. Is <br> B. Is not
Rasek [7]

Answer:

a

Explanation:

it is an electrolyte because of its strong polar chemical bond

8 0
3 years ago
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Calculate the volume of liquid in the tank sketched below. Give your answer in liters, and round to the nearest 0.1 L.
wariber [46]

Answer:

1.3 L

Explanation:

The volume of a rectangular cube can be calculated using the following formula:

Volume (L) = length (cm) x width (cm) x height (cm)

Keep in mind that 1 L = 1,000 cm³.

Before you can plug the values into the equation, you need to make sure they all have the same unit. Since the length is in meters (m), you need to first convert it to centimeters.

1 meter = 100 cm

 0.159 m           100 cm
---------------  x  ----------------  =  15.9 cm
                           1 m

Now, you can solve for the volume. To find the answer is the unit liters, you need to divide the volume by 1,000.

Volume = l x w x h

Volume = 15.9 cm x 10.5 cm x 7.7 cm

Volume = 1,285.5 cm³

Volume = 1.2855 L ------>  Volume = 1.3 L

7 0
2 years ago
How is chemical and physical properties of matter like in different
Tasya [4]
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4 0
3 years ago
The equation for the combustion of CH4 (the main component of natural gas) is
Lera25 [3.4K]

Heat produced =  -13588.956 kJ

<h3>Further explanation</h3>

Given

The reaction of combustion of Methane

CH4(g)+2O2(g)→CO2(g)+2H2O(g) ΔH∘rxn=−802.3kJ

271 g of CH4

Required

Heat produced

Solution

mol of 271 g CH₄ (MW=16 g/mol0

mol = mass : MW

mol = 271 : 16

mol = 16.9375

So Heat produced :

= mol x ΔH°rxn

= 16.9375 mol x −802.3kJ/mol = -13588.956 kJ

6 0
3 years ago
Using the balanced equation for the combustion of ethane: 2C2H6 + 7O2 → 4CO2 + 6H2O, how many moles of O2 needed to produce 12 m
pickupchik [31]

Answer:

14 moles of oxygen needed to produce 12 moles of H2O.

Explanation:

We are given that balance eqaution

2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O

We have to find number of moles of O2 needed  to produce 12 moles of H2O.

From given equation

We can see that

6 moles of   H2O produced by Oxygen =7 moles

1 mole of   H2O produced by Oxygen=\frac{7}{6}moles

12 moles of H2O produced by Oxygen=\frac{7}{6}\times 12moles

12 moles of H2O produced by Oxygen=7\times 2moles

12 moles of H2O produced by Oxygen=14 moles

Hence, 14 moles of oxygen needed to produce 12 moles of H2O.

3 0
3 years ago
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