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4 years ago

Please help me!! PLEASE

Chemistry

1 answer:

Irina18 [472]4 years ago

8 0

Answer:

synthesis

Explanation:

accorrding to wikipedia

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When 220 mL of 1.50x10^-4 M hydrochloric acid is added to 135 mL of 1.75x10^-4 M Mg(OH)2, the resulting solution will be________

siniylev [52]

Answer:

Basic

Explanation:

Considering

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

So,

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

For hydrochloric acid :

Molarity = $1.50\times 10^{-4}$ M

Volume = 220 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 220×10⁻³ L

$Moles =1.50\times 10^{-4} \times {220\times 10^{-3}}\ moles$

Moles of hydrochloric acid = 0.000033 moles

For $Mg(OH)_2$ :

Molarity = $1.75\times 10^{-4}$ M

Volume = 135 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 135×10⁻³ L

$Moles =1.75\times 10^{-4} \times {135\times 10^{-3}}\ moles$

Moles of $Mg(OH)_2$ = 0.000024 moles

According to the given reaction:

$Mg(OH)_2_{(aq)}+2HCl_{(aq)}\rightarrow MgCl_2_{(s)}+2H_2O_{(aq)}$

1 mole of $Mg(OH)_2$ reacts with 2 moles of HCl

Also,

0.000024 mole of $Mg(OH)_2$ reacts with 2*0.000024 moles of HCl

Moles of HCl = 0.000048 moles

Available moles of HCl = 0.000033 moles

Limiting reagent is the one which is present in small amount. Thus, HCl is limiting reagent.

Thus, HCL will be consumed completely and $Mg(OH)_2$ will be left over. Thus, the resulting solution will be basic.

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4 years ago

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pshichka [43]

Answer:

Explanation:

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3 years ago

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alukav5142 [94]

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3 years ago

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1-Bromopropane is treated with each of the following reagents. Draw the major substitution product if the reaction proceeds in g

sammy [17]

Answer:

Explanation:

If we look at the structure of 1-Bromopropane; we will see that it is a derivative of alkane family by the the substitution of an alkyl group. The position of the Bromine in the propane is 1, making 1-Bromopropane a primary alkyl-halide.

Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used otherwise it will result to a reactive mechanism if a weak electrophile is used.

However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.

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3 years ago

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Mademuasel [1]

Answer: Yes

Explanation:

This simulates the way offspring are formed by sperm from the male fish combining randomly with eggs from the female fish.

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3 years ago

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