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asambeis [7]
2 years ago
7

Is mendelevium flammable

Chemistry
1 answer:
Alik [6]2 years ago
6 0
No it is not flammable
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During a combustion reaction, 12.2 grams of methane reacts with 14 g of oxygen. The reaction produces carbon dioxide and water.
iogann1982 [59]

The reaction produces 6 g of carbon dioxide.

The word equation is

methane + oxygen → carbon dioxide + water

 12.2 g    +    4 g     →            x g          +  20 g

The <em>Law of Conservation of Mass</em> tells us the total mass of the reactants must equal the mass of the products. Then

12.2 g + 14 g =<em> x</em> g + 20 g

<em>x</em> = 12.2 + 14 – 20 = 6

The mass of carbon dioxide is 6 g.

<em>Note</em>: The answer can have <em>no decimal places</em> because you gave none for the masses of oxygen and water.

5 0
3 years ago
If an alkaline battery produces a cell potential of 1.46 v, what is the value of δgcell? the half-reactions in alkaline batterie
Gala2k [10]
The formula we're gonna use for this problem is written below:

ΔG°= nFE°
where
n is number of mol electrons displaced in the reaction
F is Faraday's constant = 96,500 C/mol e
E° is the standard emf

ΔG° = (2)(96,500)(1.46) = <em>281,780 Joules</em>
4 0
3 years ago
Reading the temperature of a solution by using a thermometer is an example of a(n) ________.
weeeeeb [17]

Answer:

i guess its example of observation

6 0
2 years ago
Gaseous butane, CH3(CH2)2CH, reacts with gaseous oxygen gas, O2, to produce gaseous carbon dioxide, CO2, and gaseous water, H2O.
weeeeeb [17]

Answer:

Percentage yield of carbon dioxide is 49.9%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3(CH2)2CH3 + 13O2 —> 8CO2 + 10H2O

OR

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of butane and oxygen that reacted and the mass of carbon dioxide produced from the balanced equation. This is illustrated below:

Molar mass of butane C4H10 = (12×4) + (10×1)

= 48 + 10

= 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen to produce 352 g of carbon dioxide.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen.

Therefore, 34.29 g of butane will react with = (34.29 × 416) / 116 = 122.97 g of oxygen.

From the calculation made above, we can see clearly that only 122.97 g out of 165.7 g of oxygen reacted completely with 34.29 g of butane. Therefore, butane is the limiting reactant and oxygen is the excess reactant.

Next, we shall determine the theoretical yield of carbon dioxide.

In this case, we shall use the limiting reactant because it will give the maximum yield of carbon dioxide as all of it is used up in the reaction.

The limiting reactant is butane and the theoretical yield of carbon dioxide can be obtained as follow:

From the balanced equation above,

116 g of butane reacted to produce 352 g of carbon dioxide.

Therefore, 34.29 g of butane will react to produce = (34.29 × 352) / 116 = 104.05 g of carbon dioxide.

Therefore, the theoretical yield of carbon dioxide is 104.05 g

Finally, we shall determine the percentage yield of carbon dioxide as follow:

Actual yield of carbon dioxide = 51.9 g

Theoretical yield of carbon dioxide = 104.05 g

Percentage yield of carbon dioxide =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of carbon dioxide = 51.9 / 104.05 × 100

Percentage yield of carbon dioxide = 49.9%

7 0
3 years ago
How many liters of N2 gas is needed to produce 500 L NH3?
Veseljchak [2.6K]
I think the answer is 101.2 L
6 0
3 years ago
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