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Alik [6]
3 years ago
15

What type of reaction does this equation represent? 2I4O9(s) → 4I2(s) + 9O2(g)

Chemistry
1 answer:
Tju [1.3M]3 years ago
4 0

Answer:

2I4O9_4I +9O

Explanation:

Complex compound (

I

4

O

9

) is broken down to form

I

2

and

O

2

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44.90 g of reactants are placed in a beaker
sergiy2304 [10]

Answer:

D. -1882J

Explanation:

We can solve the energy released in a chemical reaction in an aqueous medium using the equation:

Q = -m*C*ΔT

<em>Where Q is energy (In J),</em>

<em>m is mass of water (45.00g)</em>

<em>C is specific heat of water (4.184J/g°C)</em>

<em>And ΔT is change in temperature (25.00°C - 15.00°C = 10.00°C)</em>

<em />

Replacing:

Q = -45.00*4.184J/g°C*10.00°C

Q = -1882J

Right answer is:

<h3>D. -1882J</h3>

<em />

7 0
2 years ago
The total bonding energy for the products of a reaction is 2535 kJ/mol and the bonding energy of the reactants is 1375 kJ/mol. C
omeli [17]

Answer:

1160 kJ/mol

Explanation:

To get the enthalpy of the reaction take ΔH products- ΔH reactants

5 0
4 years ago
2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

6 0
3 years ago
How is a gamete different from other types of cells in a person's body?
vagabundo [1.1K]
The human gametes differ from all the other cells in the body because the somatic cells have a complete set of chromosomes
6 0
3 years ago
Use the formation reactions below such that when added together, they match the balanced equation for the combustion of methane.
muminat

Answer:

ΔH of the reaction is -802.3kJ.

Explanation:

Using Hess's law, you can know ΔH of reaction by the sum of ΔH's of half-reactions.

Using the reactions:

<em>(1) </em>Cgraphite(s)+ 2H₂(g) → CH₄(g) ΔH₁ = −74.80kJ

<em>(2) </em>Cgraphite(s)+ O₂(g) → CO₂(g) ΔH₂ = −393.5k J

<em>(3) </em>H₂(g) + 1/2 O₂(g) → H₂O(g) ΔH₃ = −241.80kJ

The sum of (2) - (1) produce:

CH₄(g) + O₂(g) → CO₂(g) + 2H₂(g) ΔH' = -393.5kJ - (-74.80kJ) = -318.7kJ

And the sum of this reaction with 2×(3) produce:

CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(g) And ΔH = -318.7kJ + 2×(-241.80kJ) =

<em>-802.3kJ</em>

7 0
3 years ago
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