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Vladimir79 [104]
2 years ago
9

Methane gas (ch4), on complete combustion in air, produces: 1. co2 2. h2 3. h2o methane gas (ch4), on complete combustion in air

, produces: 1. co2 2. h2 3. h2o 1 and 2 only 2 and 3 only 1 and 3 only all of 1 and 2 only none of
Chemistry
1 answer:
insens350 [35]2 years ago
5 0
Combustion equation of Methane:
CH₄ + 2 O₂ → CO₂ + 2 H₂O

So from the equation we can see that methane gas (CH₄) burned in air to give carbon dioxide and water

The provided choices are 1. CO₂         2. H₂         3. H₂O 

So from the choices the correct answer will be 1 and 3 only
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Acetone is one of the most important solvents in organic chemistry. It is used to dissolve everything from fats and waxes to air
Yanka [14]

Answer: a. 79.6 s

b. 44.3 s

c. 191 s

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{8.7\times 10^{-3}s^{-1}}=79.6s

b) for completion of 32% of reaction  

t=\frac{2.303}{k}\log\frac{100}{100-32}

t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{68}

t=44.3s

c) for completion of 81 % of reaction  

t=\frac{2.303}{k}\log\frac{100}{100-81}

t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{19}

t=191s

4 0
1 year ago
In an experiment, hydrochloric acid reacted with different volumes of sodium thiosulfate in water. A yellow precipitate was form
iris [78.8K]

Answer:

I think that the trend that would be seen in the time column of the data table would be that the number of seconds would increase. I know this because for each flask, the concentration of sodium thiosulfate decreases, since less of it is being mixed with more water. Also, when the concentration of a substance decreases, then the reaction rate also decreases, as there will be fewer collisions with sulfuric acid if there are fewer moles of sodium thiosulfate. When there are fewer collisions in a reaction, the reaction itself will take longer, and so when the sodium thiosulfate is diluted, the reaction takes more time.

Explanation:

<em>I verify this is correct. </em>

6 0
1 year ago
What is the molarity of a solution that contains 2.35 g of nh3 in 0.0500 l of solution?
tankabanditka [31]
The molarity is the number of moles in 1 L of the solution. 
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol 
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M
3 0
1 year ago
Read 2 more answers
If 15.6 g of hydrate are heated and only 11.7 g of anhydrous salt remain, calculate the % of water lost.
Elodia [21]

Answer:

Percent loss of water = 25%

Explanation:

Given data:

Mass of hydrated salt = 15.6 g

Mass of anhydrous salt = 11.7 g

Percentage of water lost = ?

Solution:

First of all we will calculate the mass of water in hydrated salt.

Mass of water =  Mass of hydrated salt - Mass of anhydrous salt

Mass of water = 15.6 g - 11.7 g

Mass of water = 3.9 g

Now we will calculate the percentage.

Percent loss of water = mass of water / total mass × 100

Percent loss of water = 3.9 g/ 15.6 g × 100

Percent loss of water = 25%

8 0
2 years ago
A 2400.-gram sample of an aqueous solution contains 0.012 gram of nh3. What is the concentration of nh3 in the solution
Kryger [21]

Answer : The concentration of NH_3 in the solution is, 2.94\times 10^{-4}M

Explanation :

First we have to calculate the volume of aqueous solution that is water.

Density of water = 1.00 g/mL

Mass of water = 2400 g

\text{Volume of water}=\frac{Mass}{Density}=\frac{2400g}{1.00g/mL}=2400mL

Now we have to calculate the concentration of ammonia solution.

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Mass of }NH_3\times 1000}{\text{Molar mass of }NH_3\times \text{Volume of solution (in mL)}}

Molar mass of NH_3 = 17 g/mole

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{0.012g\times 1000}{17g/mole\times 2400mL}=3.12mole/L=2.94\times 10^{-4}M

Therefore, the concentration of NH_3 in the solution is, 2.94\times 10^{-4}M

7 0
2 years ago
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