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2 years ago

Please help me pleaseeeeeeeeeeeeeeeeeeeee

Chemistry

2 answers:

Sliva [168]2 years ago

6 0

Answer:

1. false

2.c. axis

Explanation:

I love science :)

carry on learning

GaryK [48]2 years ago

3 0

Answer:

C. axis...............

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Explain how lone pairs of electrons influence the shape of a molecule.

Ronch [10]

Explanation:

the lone pairs will be negatively charged. these have a repulsion effect on other negatively charged electrons in the shells of atoms. picture a water molecule: the lone electron pair on the top of the oxygen will have a repulsion force on the 2 hydrogen atom's orbiting electrons to cause a bent molecular geometry.

3 0

2 years ago

CH3OH can be synthesized by the reaction:

Leviafan [203]

Answer:

Explanation:

8 0

2 years ago

The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A

s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $m^3 /d$

$= 80 \times 10^5 \ I/d$

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L$

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d$

If 60 mg of alum $[Al_2(SO_4)_3.14 H_2O]$ required for one litre of the water treatment.

So Alum required for $80 \times 10^5 \ I/d$

$= 80 \times 15^5 \ I/d \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d$

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $= 60 \times 0.234 \text{ of alum ppt. per litre}$

$= 14.04$ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

= 112.32 kg/d ppt of alum

Daily total solid load is $= 144 \ kg/d + 112.32 \ kg/d$

= 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234$

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-}$

After addition, the aluminium hydroxide pH of water will increase due to increase in $OH^-$ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm$

And the pH is reduced into the range of 5.9 to 6.4

6 0

2 years ago

A rigid vessel contains 3.98 kg of refrigerant-134a at 700 kPa and 60°C. Determine the volume of the vessel and the total intern

Arada [10]

<u>Answer:</u> The volume of the vessel is $0.1542m^3$ and total internal energy is 162.0 kJ.

<u>Explanation:</u>

To calculate the volume of water, we use the equation given by ideal gas, which is:

$PV=nRT$

or,

$PV=\frac{m}{M}RT$

where,

P = pressure of container = 700 kPa

V = volume of container = ? L

m = Given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000 g)

M = Molar mass of R-134a = 102.03 g/mol

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of container = $60^oC=[60+273]K=333K$

Putting values in above equation, we get:

$700kPa\times V=\frac{3980g}{102.03g/mol}\times 8.31\text{L kPa }\times 333K\\\\V=154.21L$

Converting this value into $m^3$ , we use the conversion factor:

$1m^3=1000L$

So, $\Rightarrow (\frac{1m^3}{1000L})\times 154.21L$

$\Rightarrow 0.1542m^3$

To calculate the internal energy, we use the equation:

$U=\frac{3}{2}nRT$

or,

$U=\frac{3}{2}\frac{m}{M}RT$

where,

U = total internal energy

m = given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000g)

M = molar mass of R-134a = 102.03 g/mol

R = Gas constant = $8.314J/K.mol$

T = temperature = $60^oC=[60+273]K=333K$

Putting values in above equation, we get:

$U=\frac{3}{2}\times \frac{3980g}{102.03g/mol}\times 8.314J/K.mol\times 333K\\\\U=161994.6J$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, 161994.6 J = 162.0 kJ

Hence, the volume of the vessel is $0.1542m^3$ and total internal energy is 162.0 kJ.

6 0

2 years ago

Why does the flame go out on the Bunsen burner when the gas is turned off? (use the collision theory in your answer)

Gre4nikov [31]

Answer:

Explanation:

Well the gas is the fuel for the flame of course. The collision theory comes into play when the gas turns on, chemicals collide with one another. Then reactions occur causing the flame. Then when you take away the fuel, the flame stops because there is no atoms or molecules to come together/collide with one another.

Sorry if its wrong or doesn't make sense... Wish you the best of luck on whatever your doing!

8 0

3 years ago