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Musya8 [376]
3 years ago
8

Plz help!! I really really need the help as soon as possible thank pls!! Your help matters to me pls help it’s due tomorrow morn

ing and I can’t do it also I need to go to sleep and I’m tiered!❤️

Chemistry
1 answer:
Assoli18 [71]3 years ago
5 0

Answer:

6.Given,

Final Velocity =60m/s

Initial Velocity= 0

Time=10 sec

A=?

A=Final Velocity- Initial Velocity/time

=60-0/10

=60/10

=6m/s ans.

Explanation:

Acceleration = Final Velocity - Initial Velocity/Time

By using this Formula we can calculate Acceleration.

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Hydroxylamine hydrochloride is a powerful reducing agent which is used as a polymerization catalyst. It contains 5.80 mass % H,
ludmilkaskok [199]

<u>Answer:</u> The empirical formula for the given compound is H_{4}O_1N_1Cl_1=H_4NOCl

<u>Explanation:</u>

We are given:

Percentage of H = 5.80 %

Percentage of O = 23.02 %

Percentage of N = 20.16 %

Percentage of Cl = 51.02 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of H = 5.80 g

Mass of O = 23.02 g

Mass of N = 20.16 g

Mass of Cl = 51.02 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.80g}{1g/mole}=5.80moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{23.02g}{16g/mole}=1.44moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{20.16g}{14g/mole}=1.44moles

Moles of Chlorine = \frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{51.02g}{35.5g/mole}=1.44moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.44 moles.

For Hydrogen = \frac{5.80}{1.44}=4.03\approx 4

For Oxygen = \frac{1.44}{1.44}=1

For Nitrogen = \frac{1.44}{1.44}=1

For Chlorine = \frac{1.44}{1.44}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of H : O : N : Cl = 4 : 1 : 1 : 1

Hence, the empirical formula for the given compound is H_{4}O_1N_1Cl_1=H_4NOCl

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Draw structures from the following names, and determine which compounds are optically active:(c) 1,2-dibromo-2-methylbutane
dolphi86 [110]

The given compound 1,2-dibromo-2-methylbutane is an optically active compound .

Because this compound does not have plane of symmetry (POS) and center  of symmetry (COS) i.e. does not have di-symmetry . And also forms non superimposable mirror image . the compound is optically active .

It has  chiral center.

Here , the chiral carbon is attached to the 4 distinct groups such as :  methyl , ethyl , bromine , bromomethane .

<h3>What is di-symmetry?</h3>

Di-symmetry is that which have no center of symmetry and plane of symmetry and alternate axis of symmetry .

<h3>Chiral center :</h3>

Have Sp3 hybridized center (4sigma bond ) .

4 distinct group  is attached to the chiral atom. form non -superimposable mirror image .

<h3>What is optical isomerism ?</h3>

Same molecular formula and same structural formula . also have same physical and chemical properties .

They differ in their behavior towards plane polarized light (ppl) .

Learn more about chiral center here:

brainly.com/question/9522537

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These two questions please
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Answer:

1.) 3

2.) 60 CM

Explanation:

1. Density=\frac{MASS}{VOLUME}= \frac{75}{25}

2. Length*Width*Height=3*10*2

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