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In an endothermic reaction what is true of the enthalpya) is has increased
b) is has decreased
c)it has remained uncharged
d) it has at minimum been halved

Answers

Answer 1
Answer:

Answer is: a) is has increased.

There are two types of reaction:  

1) endothermic reaction (chemical reaction that absorbs more energy than it releases).

For example, the breakdown of ozone is an endothermic process. Ozone has lower energy than molecular oxygen (O₂) and oxygen atom, so ozone need energy to break bond between oxygen atoms.

2) exothermic reaction (chemical reaction that releases more energy than it absorbs).  

For example, ΔH(reaction) = -225 kJ/mol; this is exothermic reaction.

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Which of the following is an acceptable structure for 2,5,5-trimethylhept-3-yne (CH3CH2)CH(CH3)C≡CCH2CH(CH3)2 CH3CH2C(CH3)2C≡CC(CH3)3 (CH3CH2)2C(CH3)C≡CCH2CH3 (CH3CH2)C(CH3)2C≡CCH(CH3)2 (CH3CH2CH2)CH(CH3)C≡CC(CH3)3

Answers

Answer:

D. CH₃CH₂C(CH₃)₂C≡CCH(CH₃)₂  

Explanation:

You start numbering from the end closest to the triple bond (on the right). The triple bond is between C3 and C4, and there is one methyl group on C3 and two on C5.

A. CH₃CH₂CH(CH₃)C≡CCH₂CH(CH₃)₂ is wrong. The longest chain has eight C atoms, so the compound is an octyne.

B. CH₃CH₂CH(CH₃)C≡CC(CH₃)₃ is wrong. This is a molecule of 2,2,5-trimethylhept-3-yne.

C. (CH₃CH₂)₂C≡CCH₂CH₃ is wrong. This is a molecule of 6-ethyl-5-methylhept-3-yne.

E. CH₃CH₂CH₂CH(CH₃)C≡CC(CH₃)₃ is wrong. The longest chain has eight C atoms, so the compound is an octyne.

Ne ( g ) effuses at a rate that is ______ times that of Cl 2 ( g ) under the same conditions.

Answers

Answer: 1.88times as that of Cl2

Explanation:

According to Graham law of effusion , the rate of effusion is inversely proportional to the square root of the molar mass

Rate= 1/√M

R1/R2 =√M2/M1

Let the rate of diffusion of Ne= R1

And rate of diffusion of Cl2 = R2

M1 ,molar mass of Ne= 20g/mol

M2,molar mass of Cl2 =71g/mol

R1/R2 = √ (71/20)

R1/R2 = 1.88

R1= 1.88R2

Therefore the Ne effuses at rate that is 1.88times than that of Cl2 at the same condition.

In the first step, 4-sulfanilic acid reacts with sodium nitrate to form diazonium ion intermediate. Identify the Lewis acid and Lewis base in this reaction.

Answers

Answer:

In the formation of diazonium ion intermediate, the 4-sulfanilic acid acts as the Lewis acid, while the sodium nitrite is the Lewis base.

Explanation:

A Lewis acid is by definition an electron pair acceptor (such as the H+ ion, that can accept a pair of non-bonding electrons) and a Lewis base is an electron pair donor (such as the OH- ion, that can donate a pair of non-bonding electrons).

In the formation of diazonium ion intermediate, the 4-sulfanilic acid acts as the Lewis acid(by transference of a lone pair from its nitrogen atom), while the sodium nitrite is the Lewis base.

N.B its sodium nitrite, NaNO2 (which is slightly basic in solution) not nitrate NaNO3 (which is neutral in solution)

4-sulfanilic acid acts as Lewis acid, while Sodium nitrite is the Lewis base.

What are Lewis acid and Lewis base?

A Lewis acid is an electron pair acceptor (such as the H+ ion, that can accept a pair of non-bonding electrons) and a Lewis base is an electron pair donor (such as the OH- ion, that can donate a pair of non-bonding electrons).

In the formation of diazonium ion intermediate, the 4-sulfanilic acid acts as the Lewis acid(by transference of a lone pair from its nitrogen atom), while the sodium nitrite is the Lewis base.

Sodium nitrite, NaNO₂ (which is slightly basic in solution) not nitrate NaNO₃ (which is neutral in solution)

Find more information about Lewis acid and base here:

brainly.com/question/15220646

Identify which location in the periodic table you would have the
largest atomic radii.

Answers

Answer:

left to right across a period when it decreases and when it increases top to bottom in a group,

hope i helped

A generic solid, X, has a molar mass of 78.2 g/mol. In a constant-pressure calorimeter, 12.6 g of X is dissolved in 337 g of water at 23.00 °C.X(s) yeilds X(aq)The temperature of the resulting solution rises to 24.40 °C. Assume the solution has the same specific heat as water, 4.184 J/(g·°C), and that there\'s negligible heat loss to the surroundings. How much heat was absorbed by the solution?What is the enthalpy of the reaction?

Answers

Answer:

a) Q = 2047.8 J (ΔH is negative because it's an exothermic reaction)

b) ΔH = -12.7 kJ /mol

Explanation:

Step 1: Data given

Molar mass of X = 78.2 g/mol

In a constant-pressure calorimeter, 12.6 g of X is dissolved in 337 g of water at 23.00 °C.

The temperature rise to 24.40 °C

The specific heat of the solution = 4.184 J/g°C

Step 2: Calculate the total mass

Total mass of the solution is given by  

Total mass = 12.6 grams + 337 grams = 349.6 grams

Step 3: Calculate heat

Q = m*c*ΔT

⇒ m = the total mass = 349.6 grams

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

⇒ ΔT = The change of temperature = T2 - T1 = 24.40 - 23.00 = 1.40 °C

Q = 2047.8 J (ΔH is negative because it's an exothermic reaction)

What is the enthalpy of the reaction?

Calculate number of moles = mass/ molar mass

Moles X = 12.6 grams / 78.2 g/mol

Moles X = 0.161 moles

ΔH = -2047.8 J / 0.161 moles

ΔH = -12719.3 J/mol = -12.7 kJ /mol

A solution prepared by dissolving12.6 g of X in 337 g of water, whose temperature increases from 23.00 °C to 24.00 °C, absorbs 2.05 × 10³ J of heat. The enthalpy of the reaction is -12.7 kJ/mol.

We have a solution prepared by dissolving 12.6 g of X (solute) in 337 g of water (solvent). The mass of the solution (m) is:

m = 12.6g + 337 g = 350. g

The temperature of the solution increases from 23.00 °C to 24.40 °C. Assuming that the solution has the same specific heat as water (c = 4.184 J/(g·°C)), we can calculate the heat absorbed (Q) by the solution using the following expression.

Q = c * m * \Delta T = (4.184J)/(g.\° C) * 350. g * (24.40 \° C - 23.00 \° C) = 2.05 * 10^(3) J

According to the law of conservation of energy, the sum of the heat absorbed by the solution and the heat released by the reaction is zero.

Qsol + Qreaction = 0\nQreaction = -Qsol = -2.05 * 10^(3) J

The dissolution of 12.6 g of X (molar mass 78.2 g/mol) leads to the release of 2.05 × 10³ J (hence the negative sign). The enthalpy of the reaction is

\Delta H^(0) = (-2.05 * 10^(3) J)/(12.6g) * (78.2g)/(1mol) = -1.27 * 10^(4) J/mol = -12.7 kJ/mol

A solution prepared by dissolving12.6 g of X in 337 g of water, whose temperature increases from 23.00 °C to 24.00 °C, absorbs 2.05 × 10³ J of heat. The enthalpy of the reaction is -12.7 kJ/mol.

You can learn more about calorimetry here: brainly.com/question/16104165

A sample of iron having a mass of 93.3g is heated to 65.58OC is placed in 75.0g ofwater raising the temperature from 16.95 OC to 22.24 OC. Find the specific heatcapacity for this iron sample. The answer you find has had some lab errors due tohuman mistakes. Find your percent error for your work using %Error = [(Expected - Actual) / (Expected Yield)] x100

Answers

The heat gained or lost by a substance undergoing a change in temperature is:
Q = mCpΔT
The heat lost by the iron is equal to that gained by water. The Cp for water is 4.186 Joules/gram

75 x 4.186 x (22.24 - 16.95) = -93.3 x Cp x (22.24 - 65.58)
Cp = 0.411 J/g

The heat capacity of iron is 0.45 J/g

%Error = [(0.45 - 0.41) / 0.41] x 100
%Error = 9.76%
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