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2 years ago

PLEASE HELP

Chemistry

1 answer:

LenaWriter [7]2 years ago

5 0

Hello!

I don't know a lot of this but after doing just a bit of research I think the answer is either A. or C. but I think it is A.

Hope it helps?

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Click the "draw structure" button to launch the drawing utility. under certain reaction conditions, 2,3−dibromobutane reacts wit

REY [17]

Answer:

Explanation:

According to this. Let's analize the possible products a, b and c.

First, the problem states that we have 2 eq. of base, For this case, let's assume it's KOH. Now, As we are doing a reaction with base, means that this reaction can only take places under conditions of SN2 and E2, a fast reaction that is taking place in only 1 step.

With this in mind, let's analyze product a. This states that it has two sp hybridized carbon, in other words, a triple bond between two carbons. So the product is with no doubt, an alkyne.

Product b has only one sp hybridized carbon, which means that this carbon should cannot be an alkyne because we need two carbon atoms. The only way to have one atom of C sp hybridized, is with two double bonds, so product b would have to a alkene with two double bonds.

Product c do not have sp hybridized carbon, therefore, it only has two double bonds in two different Carbon atoms, so it's another alkene with two double bonds, but in two different atoms of carbon.

Picture attached show the product a, b and c. Hope this can help

4 0

2 years ago

1 loop in the primary coil and 8 loops in the secondary. If the secondary voltage is 120 V, what must be the primary voltage

Jet001 [13]

Answer:

$V_{p} = 15 V$

Explanation:

Given the following data;

Number of loops in primary coil, Np = 1 loop.

Number of loops in secondary coil, Ns = 8 loops

Voltage in secondary coil, Vs = 120V

To find the voltage in the primary coil, Vp;

Transformer ratio is given by the formula;

$\frac{V_{p}}{V_{s}} = \frac{N_{p}}{N_{s}}$

Making Vp the subject of formula;

$V_{p} = \frac{N_{p}}{N_{s}} * V_{s}$

Substituting into the equation, we have;

$V_{p} = \frac{1}{8} * 120$

$V_{p} = \frac{120}{8}$

$V_{p} = 15 V$

Therefore, the voltage in the primary coil, Vp is 15 Volts.

4 0

2 years ago

How many significant figures are in 5.40

kotykmax [81]

There are 3 significant figures in this value, all values before and after the decimal point are significant. As there is a decimal point, the zeros trailing are also significant.

4 0

2 years ago

rite stepwise equations for protonation or deprotonation of this polyprotic species in water. What are the products of the first

adoni [48]

Answer: The products of the given chemical equation are $HCO_3^-\text{ and }OH^-$

Explanation:

Protonation equation is defined as the equation in which protons get added in the substance.

The chemical equation for the protonation of carbonate ion in the presence of water follows:

$CO_3^{2-}+H_2O\rightarrow HCO_3^-+OH^-$

By Stoichiometry of the reaction:

1 mole of carbonate ion reacts with 1 mole of water to produce 1 mole of hydrogen carbonate ion and 1 mole of hydroxide ion

Hence, the products of the given chemical equation are $HCO_3^-\text{ and }OH^-$

3 0

3 years ago

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

Answer: The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

Explanation:

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

2 years ago