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AysviL [449]
3 years ago
7

An empty fuel tank can still contain ________ and therefore can be even more dangerous than one full of liquid fuel.

Chemistry
2 answers:
Serggg [28]3 years ago
8 0

Answer:

Fumes / vapors.

Explanation:

An empty tank appears to be non dangerous as it has no liquid fuel. However even if tank is empty there may be some drops of left over liquid fuel.

These drops make the container filled with dangerous vapors that are more prone to catch fire as compared to gasoline.

Even a small spark may cause a severe explosion.

alexgriva [62]3 years ago
5 0

Answer:

An empty fuel tank can still contain "fumes"

Explanation:

Even if there is not enough liquid fuel, whatever is left in the tank creates fumes, which is more combustible than liquid gas.

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Brainliest for correct answer please show all work
Korvikt [17]

Answer:

1) Na₃PO₄ + 3 KOH ➙ 3 NaOH + K₃PO₄

2) MgF₂ + Li₂CO₃➙ MgCO₃ + 2 LiF

3) P₄ + 3 O₂➙ 2 P₂O₃

Explanation:

To balance an equation, ensure that the number of atoms of each element is equal on both sides.

Reactants would be those on the left of the arrow while products are on the right of the arrow.

Balance O and H atoms last.

<u>Question 1:</u>

__Na₃PO₄ + __KOH ➙ __NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 1Na, 1P, 3K, 5O, 1H

<em>Balance the number of Na:</em>

__Na₃PO₄ + __KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 3Na, 1P, 3K, 7O, 3H

<em>Balance the number of K:</em>

__Na₃PO₄ + 3 KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 3K, 7O, 3H

Products: 3Na, 1P, 3K, 7O, 3H

<em>The equation is now balanced.</em>

<u>Question 2:</u>

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + __LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 1F, 1Li, 1C, 3O

<em>Balance</em><em> </em><em>n</em><em>u</em><em>m</em><em>b</em><em>e</em><em>r</em><em> </em><em>of</em><em> </em><em>L</em><em>i</em><em> </em><em>and</em><em> </em><em>F</em><em> </em><em>atoms</em><em>:</em>

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + 2 LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 2F, 2Li, 1C, 3O

<em>The</em><em> </em><em>equation</em><em> </em><em>is</em><em> </em><em>now</em><em> </em><em>balanced</em><em>.</em>

<u>Question 3:</u>

__P₄ + __O₂➙ __P₂O₃

Reactants: 4P, 2O

Products: 2P, 3O

<em>Balance</em><em> </em><em>the</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>P</em><em> </em><em>atoms</em><em>:</em>

__P₄ + __O₂➙ 2 P₂O₃

Reactants: 4P, 2O

Products: 4P, 6O

<em>Balance</em><em> </em><em>the</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>O</em><em> </em><em>atoms</em><em>:</em>

__P₄ + 3 O₂➙ 2 P₂O₃

Reactants: 4P, 6O

Products: 4P, 6O

<em>The</em><em> </em><em>equation</em><em> </em><em>is</em><em> </em><em>now</em><em> </em><em>balanced</em><em>.</em>

4 0
3 years ago
the average single and double n-o bond lengths are 136 pm and 122 pm, respectively. assuming these two structures were to make e
uranmaximum [27]

Resonance structures show that chemical species have equal bond lengths and bond angles when in resonance. The bond lengths of the N - O bond in HNO2 is 126pm.

Linus Pauling introduced the idea of resonance to explain the nature of bonding in compounds and ions where a single Lewis structure can not satisfactorily account for the observed properties of the chemical species.

If the two structures are resonance structures, they will have equal N - O bond lengths and bond angles. The bond length will be intermediate between that of a pure N - O single and N - O double bond. The actual N- O bond lengths in HNO2 are obtained as;  136 pm -  122 pm/2 = 126 pm.

Learn more: brainly.com/question/8155254

6 0
2 years ago
11. A sample contains 25% water and weighs 201 grams. Determine the grams of water in the
77julia77 [94]

Answer:

m_{water}=50.25g

Explanation:

Hello,

In this case, considering that the by-mass percent of water is:

\% m/m=\frac{m_{water}}{m_{sample}}*100\%

Given such percent and the mass of the sample, we can find the mass of water in grams in the sample by solving for it as shown below:

m_{water}=\frac{\%m/m*m_{sample}}{100\%}\\ \\m_{water}=\frac{25\%*201g}{100\%}\\ \\m_{water}=50.25g

Best regards.

5 0
3 years ago
2KCIO3 --- 2 KCI + 3O2
Juli2301 [7.4K]

Answer:

465.69g KCIO3

Explanation:

See the stoichiometry in the reaction:

We can propose

3 moles of oxgen are made of 2 moles of chlorate

Therefore 5.70 moles of O₂ will be made by (5.70. 2) / 3 = 3.8 moles of chlorate.

We convert the moles to mass: 3.8 mol . 122.55 g/ 1 mol = 465.69g KCIO3

is the mass to use in the begining

3 0
3 years ago
Read 2 more answers
What happens to a catalyst used in a reaction?
scoundrel [369]

Answer:

It is broken up by the reactants

4 0
2 years ago
Read 2 more answers
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