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3 years ago

In what type of reaction or process does heat flow into the system

Chemistry

1 answer:

SCORPION-xisa [38]3 years ago

5 0

Answer: The correct answer is: " endothermic . "

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Note: Heat flows into [heat may be absorbed within] an "endothermic" reaction or system

To the contrary, heat flows out  [heat may exit from or may be released from] an "exothermic" reaction or process.

Hint: Think of the "prefixes" of: "endothermic" and "exothermic" :

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1) endo- = "within" (as in "endothermic" —heat tends to be absorbed/"within"/"released within"/released within"/into" ;

2) exo- = " outwards"/"exit" (as in "exothermic") —heat tends to '"exit"/leave/escape from/"be released out of/form".

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Hope this is helpful to you!

Best wishes to you in your academic pursuits

—and within the "Brainly" community"!

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You might be interested in

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

5 0

3 years ago

) Consider the starting materials and reagents. What do you expect to happen at the beginning of this reaction (Boxes 1-3 on the

tekilochka [14]

Answer:

Check the explanation

Explanation:

functional group found in the major organic product = alpha -beta unsaturated ketone

Reaction used to form this functional group = Michael condensation reaction

Also other reactions are - Aldol condensation , Robinson annulation reaction.

Kindly check the attached image below to see the step by step solution to the question above.

3 0

3 years ago

NiS2(s) + O2(g) --> NiO(s) + SO2(g) When 11.2 g of NiS2 react with 5.43 g of O2, 4.86 g of NiO are obtained. The theoretical

makkiz [27]

Answer:

1. The theoretical yield of NiO is 5.09g.

2. O2 is the limiting reactant.

3. The percentage yield of NiO is 95.5%

Explanation:

Step 1:

The balanced equation for the reaction is given below:

2NiS2(s) + 5O2(g) —> 2NiO(s) + 4SO2(g)

Step 2:

Determination of the masses of NiS2 and O2 that reacted and the mass of NiO produced from the balanced equation. This is illustrated below below:

Molar mass of NiS2 = 59 + (32x2) = 123g/mol

Mass of NiS2 from the balanced equation = 2 x 123 = 246g

Molar mass of o3= 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g

Molar mass of NiO = 59 + 16 = 75g/mol

Mass of NiO from the balanced equation = 2 x 75 = 150g

Summary:

From the balanced equation above, 246g of NiS2 reacted with 160g of O2 to produce 150g of NiO

Step 3:

Determination of the limiting reactant. This can be obtain as follow:

From the balanced equation above, 246g of NiS2 reacted with 160g of O2.

Therefore, 11.2g of NiS2 will react with = (11.2 x 160)/246 = 7.28g of O2.

From the above calculation, we can see that it will take a higher mass of O2 i.e 7.28g than what was given i.e 5.43g to react completely with 11.2g of NiS2.

Therefore, O2 is the limiting reactant and NiS2 is the excess reactant.

1. Determination of the theoretical yield of NiO.

In this case, the limiting reactant will be used as all of it is consumed in the reaction. The limiting reactant is O2.

From the balanced equation above, 160g of O2 reacted to produce 150g of NiO.

Therefore, 5.43g of O2 will react to produce = (5.43 x 150)/160 = 5.09g of NiO.

Therefore, the theoretical yield of NiO is 5.09g.

2. The limiting reactant is O2. Please review step 3 above for explanation.

3. Determination of the percentage yield of NiO. This is illustrated below:

Actual yield of NiO = 4.86g

Theoretical yield of NiO = 5.09g

Percentage yield =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 4.86/5.09 x 100

Percentage yield of NiO = 95.5%

3 0

3 years ago

You have 15.0 L of nitrogen gas at 100°C and 200 kPa. What volume of gas will you have at STP?

Pavel [41]

Actually, you do NOT need to calculate the number of moles, so you do NOT need to convert the volume and pressure. (You DO have to convert the temperature.) That's because for this problem, pV/T is constant, so the units cancel (but the temperature scale matters).

200kPa * 15L / 373K = 101kPa * V / 273K 
V = 21.7 L

7 0

4 years ago

An aqueous solution 10 g of an optically pure substance diluted to 500ml with water and placed in a polarimeter tube 20 cm long.

Gennadij [26K]

Explanation:

Formula to calculate specific rotation is as follows.

Specific rotation ( $[\alpha]$ ) = $\frac{\alpha}{c} \times l$