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2 years ago

4 possible careers in chemistry and place of application

Chemistry

1 answer:

LenaWriter [7]2 years ago

5 0

Answer:

Analytical Chemist.

Chemical Engineer.

Chemistry Teacher.

Forensic Scientist.

<em>PLZ MARK THE BRAINLIST</em>

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If a solution has a pH of 7.41, determine its H+ concentration.

makvit [3.9K]

$[H+]=10^-^p^H$

$[H+]=10^-^7^.^4^1$

$[H+] = 3.89$ $x$ $10^-^8$

8 0

3 years ago

If 45.0 mL of ethanol (density = 0.789 g/mL) initially at 8.0 ∘C is mixed with 45.0 mL of water (density = 1.0 g/mL) initially a

strojnjashka [21]

Answer : The final temperature of the mixture is, $22.14^oC$

Explanation :

First we have to calculate the mass of ethanol and water.

$\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}=0.789g/mL\times 45.0mL=35.5g$

and,

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.0g/mL\times 45.0mL=45.0g$

Now we have to calculate the final temperature of the mixture.

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.42J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol = 35.5 g

$m_2$ = mass of water = 45.0 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $8.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

Now put all the given values in the above formula, we get:

$35.5g\times 2.42J/g^oC\times (T_f-8.0)^oC=-45.0g\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.14^oC$

Therefore, the final temperature of the mixture is, $22.14^oC$

3 0

3 years ago

A weak acid, ha, has a p��a of 4. 524. If a solution of this acid has a ph of 4. 799 , what percentage of the acid is not io

insens350 [35]

Answer:

highly endothermic

Explanation:

6 0

1 year ago

The internet is a local communication network that allows only certain computers to connect and exchange information True or fal

choli [55]

False, the internet branches throughout many countries connecting you to the rest of the world from home. Hope it helped!

6 0

2 years ago

For the equilibrium PCl5(g) PCl3(g) + Cl2(g), Kc = 2.0 × 101 at 240°C. If pure PCl5 is placed in a 1.00-L container and allowed

Cloud [144]

Answer:

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

Explanation:

for the reaction

PCl₅(g) → PCl₃(g) + Cl₂(g)

where

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 2.0*10¹ M = 20 M

and [A] denote concentrations of A

if initially the mixture is pure PCl₅ , then it will dissociate according to the reaction and since always one mole of PCl₃(g) is generated with one mole of Cl₂(g) , the total number of moles of both at the end is the same → they have the same concentration → [PCl₃(g)] = [Cl₂]=0.27 M

therefore

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 0.27 M* 0.27 M /[PCl₅] = 20 M

[PCl₅] = 0.27 M* 0.27 M / 20 M = 3.64*10⁻³ M

[PCl₅] = 3.64*10⁻³ M

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

6 0

3 years ago