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3 years ago

What would likely happen if you were to touch the flask in which an endothermic reaction occurring

Chemistry

2 answers:

coldgirl [10]3 years ago

5 0

Answer:

The reaction vessel will feel cool

Explanation:

In an endothermic reaction, heat is absorbed by the reaction system from the surrounding. The withdrawal of heat from the surrounding causes the reaction vessel to feel cool.

Hence when an endothermic reaction is going on, the reaction vessel feels cool when touched.

Black_prince [1.1K]3 years ago

3 0

Answer:

The flask would probably feel cooler than before the reaction started.

Explanation:

An endothermic reaction means that it needs heat to continue. The required heat will be taken from the flask and by touching it the reaction reactants will cause the solution contained in the flask to cool with a drop in temperature.

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2 years ago

Some properties of matter depend on how much matter there is. Put an X next to all of the statements you think are true about pr

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Answer:

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Explanation:

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3 years ago

How many molecules are in 92 liters of butane at STP<br><br> please show work

natita [175]

2.5 x 10^24 molecules of C4H10

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3 years ago

A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo

atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

Weight = $0.09 \times 16.04 + 0.91 \times 29$

= 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

$\frac{800}{27.8336}$

= 28.74 kmol/hr

Now, applying component balance as follows.

$0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}$

$F_{p}$ = 43.11 kmol/hr

$F_{A} + F_{B} = F_{p}$

$F_{B}$ = 43.11 - 28.74

= 14.37 kmol/hr

So, mass flow rate of pure (B), is $F_{B}$ = $14.37 \times 29$

= 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

Molecular weight of product stream = $Mol. weight \times 43.11 kmol/hr$

= $0.06 \times 16.04 + 0.94 \times 29$

= 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of $O_{2}$ into the product stream is as follows.

$0.21 \times 0.94 \times 43.11$

= $8.5099 kmol/hr \times 329 g/mol$

= 272.31 kg/hr

Therefore, calculate the mass % of $O_{2}$ into the stream as follows.

$\frac{272.31}{1216.67} \times 100$

= 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of $O_{2}$ in the product gas is 22.38%.

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3 years ago

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NNADVOKAT [17]

if the compound is made of just two elements, if one is a metal (ie belongs to any of groups 1, 2 or 3) and the other element a non metal, (ie belongs to group 5, 6 or 7) then the compound is most likely to be an ionic compound. For example NaCl, MgO
If the compound is made of identical non metalic elements as in O2, Cl2 then the compound is covalent
If the compoud is made of just two elements that are both non metals such as in SO2, CO, NO, CCl4, the compound is covalent
If the compound is made up of more than two elements, such as in HNO3, Na2CO3, CuSO4.5H2O, you may need to break the compound into dissociating parts. You will see that, the compounds are ionic.
Hydrocarbons, compounds containing only hydrogen and carbon of varying molecular size are all covalent. Examples are C2H6, C2H4, C2H2

Note that there could be some little exceptions to the examples given. Mostly with first members of every group because of their small size which make them show substantial deviations from group behavior. For example HCl is covalent not ionic.

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3 years ago