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Ne4ueva [31]
3 years ago
13

Help me with questions 1-8. here's the link to the textbook

Chemistry
1 answer:
Ede4ka [16]3 years ago
6 0

Answer:

7 gmkrjfjrjgfhgh

Explanation:

bc I'm dum

You might be interested in
Calculate the volume of 4.00 molar NaOH solution required to prepare 100 mL of a 0.500 molar solution of NaOH.
Kazeer [188]

Answer:

The answer to your question is V₁ = 12.5 ml

Explanation:

Data

Volume = V₁?

[NaOH] = C₁ = 4.0 M

Volume 2 = V₂ = 100 ml

[NaOH] = C₂ = 0.5 M

Formula of dilution

                             V₁C₁   =  V₂C₂

Solve for V₁ (original solution)

                                 V₁ = \frac{V2C2}{C1}

Substitution

                                V₁ = \frac{(0.5)(100)}{4}

Simplification

                                V₁ = \frac{50}{4}

Result

                                V₁ = 12.5 ml

7 0
3 years ago
A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th
kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = \frac{38.8}{61.2^{2} }

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

0.0104 = \frac{200 - P(NO_{2})  }{[P(NO_{2} )]^{2}}

0.0104[P(NO_{2} )]^{2} + P(NO_{2} ) - 200 = 0

Resolving the second degree equation:

P(NO_{2} ) = \frac{-1+\sqrt{9.32} }{0.0208}

P(NO_{2} ) = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are P(NO_{2} ) = 98.7 MPa and P(N₂O₄) = 101.3 MPa

3 0
3 years ago
How many of the following species are paramagnetic? sc3+ br- mg2+ se?
Aloiza [94]
Answer:
           One: <u>Selenium</u> is Paramagnetic

Explanation:
                   Those compounds which have unpaired electrons are attracted towards magnet. This property is called as paramagnetism. Lets see why remaining are not paramagnetic.

Electronic configuration of Scandium;

Sc  =  21  = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹

Sc³⁺  =  1s², 2s², 2p⁶, 3s², 3p⁶ 

Hence in Sc³⁺ there is no unpaired electron.

Electronic configuration of Bromine;

Br  =  35  = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁵

Br⁻  =  1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶

Hence in Br⁻ there is no unpaired electron.

Electronic configuration of Magnesium;

Mg  =  12  = 1s², 2s², 2p⁶, 3s²

Mg²⁺  =  1s², 2s², 2p⁶

Hence in Mg²⁺ there is no unpaired electron.

Electronic configuration of selenium;

Se  =  34  = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁴

Or,

Se  =  34  = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4px², 4py¹, 4pz¹

Hence in Se there are two unpaired electrons hence it is paramagnetic in nature.
8 0
3 years ago
The concentration in molality of hcl in a solution that is prepared by dissolving 5.5 g of hcl in 200.0 g of c2h6o is __________
PIT_PIT [208]

Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the molality of the given solution, we need to convert the mass of solute into moles and divide it to the mass of the solvent.

<span>
Moles of HCl = 5.5 g HCl ( 1 mol HCl / 36.46 g HCl ) = 0.1509 mol HCl</span>

<span>
Molality = 0.1509 mol HCl / 200 g C2H6O ( 1 kg / 1000 g )
Molality = 0.7543 mol / kg</span>


<span>The concentration in molality of hcl in a solution that is prepared by dissolving 5.5 g of hcl in 200.0 g of c2h6o is 0.7453 molal.</span>
6 0
3 years ago
I typed a question. It said give points. I selected. Someone answered notification came. But where do I see they persons answer?
Ksivusya [100]

Answer:

212.304 grams

Explanation:

similar to your other question, use the same formula

q=mCpΔT

23617=m(4.182)(46.6-20)

23617=111.2412m

m=212.304g

6 0
2 years ago
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