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3 years ago

How many atoms are in 755 moles?

Chemistry

1 answer:

artcher [175]3 years ago

3 0

Answer:

The answer to your question is:

Explanation:

1 mol = 6.022 x 10 ²³ atoms

755 moles = x

x = 755 x 6.022 x 10 ²³ / 1 = 4.54 x 10 ²⁶ atoms

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When benzene (c6h6) reacts with bromine (br2) bromobenzene(c6h5br) is obtained: c6h6 + br2 → c6h5br + hbr what is the theoretica

saveliy_v [14]

Answer A) : We have to calculate the number of moles of Benzene involved in the reaction,

30 g / 78 moles of benzene = 0.384 moles

For bromine it will be the same process,

65 g / 159.8 moles = 0.406 moles

By observing the reaction given above we can say that the reaction ratio of bromine and benzene is 1 : 1

We need to find the mass of bromobenzene,

which should be, 6(12) + 5 (1)+ 79.90 = 156.9 g/mol

So, the mass of bromobenzene will be 156.9 g/mol X 0.3846 mol = 60.343 g

Hence the theoretical yield will be 60.34 g

Answer B) : To calculate the actual yield we have to divide it with theoretical yield.

(56.7g / 60.343 g ) X100% = 93.96 %

Here, we can say that we got 93.96 % of actual yield.

As we know it is impossible to get 100% yield in any reaction.

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3 years ago

Scientists make observations, conduct experiments, and draw conclusions. Which statement is true of all scientists?

Vsevolod [243]

Answer:

study the natural world

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3 years ago

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The formation of a white cloud

saw5 [17]

Most clouds are white. That's because water and ice particles that make up a cloud have just the right amount and sizes to scatter light in all possible wavelengths. When light of practically all wavelengths combine, the result is white light.

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3 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

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3 years ago

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liberstina [14]

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2 years ago