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Mumz [18]
3 years ago
12

Someone please help i don’t have much time left

Chemistry
1 answer:
REY [17]3 years ago
6 0

Answer: Energy of reactants = 30, Energy of products = 10

Exothermic

Activation energy for forward reaction is 10.

Explanation:

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and \Delta H for the reaction comes out to be negative.

Energy of reactants = 30

Energy of products = 10

Thus as energy of the product < energy of the reactant, the reaction is exothermic.

Activation energy (E_a) is the extra energy that must be supplied to reactants in order to cross the energy barrier and thus convert to products.

E_a for forward reaction is (40-30) = 10.

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A student measured out 0.333 moles of theobromine, CyH3N402 (MM = 180.8/mol) the
Verdich [7]

Answer

2.0 x 10²³ molecules.

Explanation

Given:

The number of moles of theobromide measured out = 0.333 moles.

MM of theobromide = 180.8 g/mol

What to find:

The number of molecules of theobromide the student measured.

To go from moles to molecules, multiply the number of moles by Avogadro's number.

The Avogadro's number = 6.02 x 10²³

1 mole of theobromide contains 6.02 x 10²³ molecules.

So, 0.333 moles of theobromide measured out will have (0.333 x 6.02 x 10²³) = 2.0 x 10²³ molecules.

8 0
1 year ago
How many moles of aluminum are in 4.35 moles of al2si4o10(OH)2
Gnoma [55]

Answer:

8.7 mol Al

Explanation:

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4.35 × 2 = 8.7

6 0
3 years ago
You add 50.0 g of ice initially at ‒20.0 °C to 1.00 x 102 mL warm water at 67.0 °C. When all the ice melts, the water temperatur
xxTIMURxx [149]

Answer:

T_2=17.8\°C

Explanation:

Hello,

In this case, we can solve this problem by noticing that the heat lost by the warm water is gained by the ice in order to melt it:

Q_{water}=-Q_{ice}

In such a way, the cooling of water corresponds to specific heat and the melting of ice to sensible heat and specific heat also that could be represented as follows:

m_{water}Cp_{water}(T_2-T_{water})=-m_{ice}\Delta H_{melting,ice}-m_{ice}\Cp_{ice}(T_2-T_{ice})

Thus, specific heat of water is 4.18 J/g°C, heat of melting is 334 J/g and specific heat of ice is 2.04 J/g°C, thus, we can compute the final temperature as shown below:

m_{water}Cp_{water}(T_2-T_{water})+m_{ice}Cp_{ice}(T_2-T_{ice})=-m_{ice}\Delta H_{melting,ice}\\\\T_2=\frac{-m_{ice}\Delta H_{melting,ice}+m_{water}Cp_{water}T_{water}+m_{ice}Cp_{ice}T_{ice}}{m_{water}Cp_{water}+m_{ice}Cp_{ice}} \\\\T_2=\frac{-50.0*334+100*4.18*67+50.0*2.04*-20.0}{100*4.18+50.0*2.04} \\\\T_2=17.8\°C

Best regards.

3 0
3 years ago
Say something hot ;)
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sometimes I think of plss, and it reminds me of you *bites lip*

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2 years ago
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