All categories

3 years ago

What else should I add for creativity? Or else I won’t receive credit

Chemistry

2 answers:

padilas [110]3 years ago

4 0

Answer:

color the star with some spacey color(purpel, lavender, white spots)

You can also cover the rest of paper with different objects from space :)

Explanation:

Oxana [17]3 years ago

4 0

Answer:

More colors and add some planets.

You might be interested in

how much hydrogen gas is necessary to exert a pressure of 1.4 ATM at 430k if occupying a volume of 15.1 l

lora16 [44]

Answer:

Explanation:

7 0

3 years ago

If Maria winks exactly 5 times every minute and she sleeps

Elina [12.6K]

First, we need to find the amount of winks in an hour, to simplify it so we can answer easier. Since there are 60 minutes in an hour, we can multiply 5 by 60 to find the number of winks in an hour.

5*60=300

Now multiply 300 by 8.

300*8=2,400

She winks exactly 2,400 times per day.

Hope this helps!

4 0

3 years ago

Read 2 more answers

Sodium (Na) and potassium (K) are in the same group in the periodic table. Sodium is in the 11th position. How many valence elec

lapo4ka [179]

It's 1 because there is only one electron on the outer shell.

8 0

3 years ago

Read 2 more answers

In one step in the synthesis of the insecticide Sevin, naphthol reacts with phosgene as shown.

Serhud [2]

Answer:

Explanation:

Chemical equation:

C₁₀H₈O + COCl₂ → C₁₁H₇O₂Cl + HCl

A. How many kilograms of C₁₁H₇O₂Cl form from 2.5×10*2 kg of naphthol?

Given data:

Mass of naphthol = 2.5 ×10² kg ( 250×1000 = 250000 g)

Mass of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 250000 g/ 144.17 g/mol

Number of moles of naphthol = 1734.1 mol

Now we will compare the moles of naphthol with C₁₁H₇O₂Cl.

C₁₀H₈O : C₁₁H₇O₂Cl

1 : 1

1734.1 : 1734.1

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass = 1734.1 mol × 206.5 g/mol

Mass = 358091.65 g

Gram to kilogram:

1 kg×358091.65 g/ 1000 g = 358.1 kg

B. If 100. g of naphthol and 100. g of phosgene react, what is the theoretical yield of C11H7O2Cl?

Given data:

Mass of naphthol = 100 g

Mass of COCl₂ = 100 g

Theoretical yield of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 100 g/ 144.17 g/mol

Number of moles of naphthol = 0.694 mol

Number of moles of phosgene:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 99 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of naphthol and phosgene with C₁₁H₇O₂Cl.

C₁₀H₈O : C₁₁H₇O₂Cl

1 : 1

0.694 : 0.694

COCl₂ : C₁₁H₇O₂Cl

1 : 1

1.0 : 1.0

The number of moles of C₁₁H₇O₂Cl produced by C₁₀H₈O are less so it will limiting reactant and limit the yield of C₁₁H₇O₂Cl.

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass = 0.694 mol × 206.5 g/mol

Mass = 143.3 g

Theoretical yield = 143.3 g

C. If the actual yield of C11H7O2Cl in part b is 118 g, what is the percent yield?

Given data:

Actual yield of C₁₁H₇O₂Cl = 118 g

Theoretical yield = 143.3 g

Percent yield = ?

Solution:

Formula :

Percent yield = actual yield / theoretical yield × 100

Now we will put the values in formula.

Percent yield = 118 g/ 143.3 g × 100

Percent yield = 0.82 × 100

Percent yield = 82%

5 0

3 years ago

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.