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gladu [14]
2 years ago
13

Charlie is frying an egg in a pan located over a gas burner. He develops a model to determine the energy produced by the flame i

n the gas burner by calculating the energy absorbed by the egg. Which assumption will Charlie need to make in order for his model to be considered a closed system?
Chemistry
1 answer:
frosja888 [35]2 years ago
8 0

Answer: B.)

Explanation:  Heat flows from pan to sounroundings

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Draw the structure of a compound with molecular formula c5h12 that exhibits only one kind of proton (all 12 protons are chemical
Sophie [7]
The middle carbon is 4-degree since it is attached to 4 carbons. All other carbons are 1-degree since they are attached to only 1 carbon. 

Hydrogens attached with 1-degree carbon are all same. Hydrogen are often refereed to as protons. No carbon is attached to 4-degree carbon. So all hydrogens in this structure are same.

This structure is called  NeoPentane

4 0
2 years ago
Website
vredina [299]

Answer:

volume in L = 0.25 L

Explanation:

Given data:

Mass of Cu(NO₃)₂ = 2.43 g

Volume of KI = ?

Solution:

Balanced chemical equation:

2Cu(NO₃)₂  + 4KI    →    2CuI + I₂ + 4KNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

Number of moles = 2.43 g/ 187.56 g/mol

Number of moles = 0.013 mol

Now we will compare the moles of Cu(NO₃)₂ with KI.

                        Cu(NO₃)₂       :              KI    

                              2              :               4

                            0.013          :            4 × 0.013=0.052 mol

Volume of KI:

<em>Molarity = moles of solute / volume in L</em>

volume in L = moles of solute /Molarity

volume in L =  0.052 mol / 0.209 mol/L

volume in L = 0.25 L

6 0
2 years ago
A ground state hydrogen atom absorbs a photon of light having a wavelength of 93.7 nm.93.7 nm. What is the final state of the hy
sammy [17]

Answer:

5

Explanation:

Given that the formula is;

1/λ= R(1/nf^2 - 1/ni^2)

λ = 93.7 nm or 93.7 * 10^-9 m

R= 1.097 * 10^7 m-1

nf = ?

ni = 1

From;

ΔE = hc/λ

ΔE = 6.63 * 10^-34 * 3* 10^8/93.7 * 10^-9

ΔE = 21 * 10^-19 J

ΔE = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)

21 * 10^-19 J = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)

21 * 10^-19/-2.18 * 10^-18 = (1/nf^2 - 1/1^2)

-0.963 = (1/nf^2 - 1)

-0.963 + 1 = 1/nf^2

0.037 = 1/nf^2

nf^2 = (0.037)^-1

nf^2 = 27

nf = 5

7 0
2 years ago
Find the molarity of 186.55 g of sugar (C12H22O11) in 250. mL of water.
Anna [14]

Answer:

The molarity of this sugar solution in water is 2.18 M

Explanation:

Step 1: Data given

Mass of sugar (C12H22O11) = 186.55 grams

Molar mass of C12H22O11 = 342.3 g/mol

Volume of water = 250.0 mL = 0.250 L

Step 2: Calculate moles sugar

Moles sugar = mass sugar / molar mass sugar

Moles sugar = 186.55 grams / 342.3 g/mol

Moles sugar = 0.545 moles

Step 3: Calculate molarity of the sugar solution

Molarity = moles sugar / volume of water

Molarity = 0.545 moles / 0.250 L

Molarity = 2.18 MThe molarity of this sugar solution in water is 2.18 M

6 0
2 years ago
A 40.0 mL sample of 0.25 M KOH is added to 60.0 mL of 0.15 M Ba(OH)2. What is the molar concentration of OH-(aq) in the resultin
solmaris [256]

Answer:

C) 0.28 M

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Moles =Molarity \times {Volume\ of\ the\ solution}

Potassium hydroxide will furnish hydroxide ions as:

KOH\rightarrow K^{+}+OH^-

Given :

<u>For Potassium hydroxide : </u>

Molarity = 0.25 M

Volume = 40.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 40.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Potassium hydroxide is same as the moles of Potassium hydroxide as shown below:

Moles =0.25 \times {40.0\times 10^{-3}}\ moles

Moles of hydroxide ions by Potassium hydroxide = 0.01 moles

Barium hydroxide will furnish hydroxide ions as:

Ba(OH)_2\rightarrow Ba^{2+}+2OH^-

Given :

<u>For Barium hydroxide : </u>

Molarity = 0.15 M

Volume = 60.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 60.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Barium hydroxide is twice the moles of Barium hydroxide as shown below:

Moles =2\times 0.15 \times {60.0\times 10^{-3}}\ moles

Moles of hydroxide ions by Barium hydroxide = 0.018 moles

Total moles = 0.01 moles + 0.018 moles = 0.028 moles

Total volume = 40.0×10⁻³ L + 60.0×10⁻³ L = 100×10⁻³ L

Concentration of hydroxide ions is:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity_{OH^-}=\frac{0.028 }{100\times 10^{-3}}

<u> The final concentration of hydroxide ion = 0.28 M</u>

5 0
2 years ago
Read 2 more answers
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