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lord [1]
3 years ago
5

What were the procedures and outcome of this experiment? Contrast the actual outcome with Rutherford’s anticipated outcome.

Chemistry
1 answer:
jarptica [38.1K]3 years ago
6 0
Be more specific, I could possibly find the asnwer you're looking towards.
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What is the product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes?
zheka24 [161]

The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .

Sodium borohydride is a relatively selective reducing agent  Ethanolic solutions of Sodium borohydride reduces aldehyde , and ketone , in the presence of acid chloride , ester , epoxide , lactones , acids , nitriles , nitro groups.

The sodium borohydride does not reduce ester group because sodium borohydride is not strong enough and the electrophilicity at carbony carbon of ester is not more as compare toaldehyde , and ketone

The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .

to learn more about sodium borohydride and ethanol click here ,

brainly.com/question/12955502

#SPJ4

4 0
2 years ago
How many chloride ions are in 2.6 moles of CaCl2?
Rama09 [41]

Answer:

31.31× 10²³ number of Cl⁻ are present in 2.6 moles of CaCl₂ .

Explanation:

Given data:

Number of moles of CaCl₂ = 2.6 mol

Number of Cl₂ ions = ?

Solution:

CaCl₂  → Ca²⁺ + 2Cl⁻

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

In one mole of CaCl₂ there are two moles of chloride ions present.

In 2.6 mol:

2.6×2 = 5.2 moles

1 mole Cl⁻ =   6.022 × 10²³ number of Cl⁻ ions

5.2 mol ×  6.022 × 10²³ number of Cl⁻  / 1mol

31.31× 10²³ number of Cl⁻

8 0
3 years ago
Base your answers on the graph below, which represents uniform cooling of a sample of a pure substance, starting as a gas. Solid
Karolina [17]

Answer:

D & E

Explanation:

I think this is dealing with latent heat and D & E would be the range where you will find solid and liquid phases in equilibrium, cuz it starts as gas at from A to B, B to C is gas and liquid equilibrium, C to D is liquid, D to E solid and liquid, and then E to F is solid.

7 0
3 years ago
PLEASE HELP!! 40 POINTS
PSYCHO15rus [73]

Answer:

1) 6.524779402×10^(-17)  

2)521.1g

3)113

Explanation:

5 0
3 years ago
Read 2 more answers
What is △n for the following equation in relating Kc to Kp?
Nimfa-mama [501]

Answer:

-1  

Explanation:

The relation between Kp and Kc is given below:

K_p= K_c\times (RT)^{\Delta n}

Where,  

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant , 0.082057 L atm.mol⁻¹K⁻¹

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

2Na_{(s)}+2H_2O_{(l)}\rightleftharpoons 2NaOH_{(aq)}+2H_2_{(g)}

<u>Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)  = (2+1)-(2+2) = -1  </u>

<u></u>

7 0
3 years ago
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