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3 years ago

When heating liquid materials in laboratory glassware, a student should always

Chemistry

2 answers:

Lelechka [254]3 years ago

5 0

Answer:

be careful to not hurt yourself

Nady [450]3 years ago

4 0

Answer: make sure the liquid is pointing away from the face

Explanation:

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How to draw Hess' Cycle for this question ?

NISA [10]

Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $C_2H_4$ will be,

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_1=-1411kJ/mole$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.7kJ/mole$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.9kJ/mole$

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :

(1) $2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H_1=+1411kJ/mole$

(2) $2C(s)+2O_2(g)\rightarrow 2CO_2(g)$ $\Delta H_2=2\times (-393.7kJ/mole)=-787.4kJ/mole$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.9kJ/mole)=-571.8kJ/mole$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+1411kJ/mole)+(-787.4kJ/mole)+(-571.8kJ/mole)$

$\Delta H=51.8kJ/mole$

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

7 0

3 years ago

500 mL of a solution contains 1000 mg of CaCl2. Molecular weight of CaCl2 is 110 g/mol. Specific gravity of the solution is 0. C

dangina [55]

Answer:

a) 0,2% w/v

b) r=500

c) 0,0182 M

d) 0,0145 m

e) 0,0137 equivalents

Explanation:

a) % w/v means mass of solute in grams per 100 mililiter of solution. Thus:

%w/v= $\frac{1,000 g CaCl2}{500mL}$ ×100 = 0,2%w/v

b) Ratio strength is a way to express concentration. For w/v is in 1g of solute r mililiters of solution have. Thus, r = 500 because we have in the first 1 g of CaCl₂ in 500 mL of solution.

c) Molarity is moles of solute per liter of solution, thus:

1,000 g of CaCl₂ × $\frac{1mol}{110g}$ = 9,09×10⁻³ moles of CaCl₂

500 mL of solution × $\frac{1L}{1000mL}$ = 0,500 L of solution

M = $\frac{9,09x10^{-3} moles }{0,500 L}$ = 0,0182 M

d) Molality is moles of solute per kg of solution.

Specific gravity is the ratio between density of the solution and density of a reference substance (Usually water). With a specific gravity of 0,8:

kg of solution = 0,500 L of solution × $\frac{0,8 kg}{1L}$ = 0,625 kg of solution

m = $\frac{9,09x10^{-3}moles }{0,625 kg}$ = 0,0145 m

e) In a salt, equivalents are the number of moles ables to replace one mole of charge. In CaCl₂ is ¹/₂ because with ¹/₂ moles of CaCl₂ it is possible to replace 1 mole of charges. Thus, in 1,5 L there are:

1,5 L × $\frac{0,0182 CaCl2 moles}{1L}$ × $\frac{1equivalent}{2 moles}$ = 0,0137 equivalents

I hope it helps!

7 0

3 years ago

Give the formula of the conjugate base of HSO4–

mr Goodwill [35]

The answer is So^2-4

7 0

3 years ago

A catalyst is used in the chemical reaction to break down hydrogen peroxide into water and oxygen. At the end of the reaction, t

Natasha_Volkova [10]

At the end of the reaction, the catalyst is UNCHANGED.
:)

8 0

3 years ago

Ch4+br2=ch3br+hbr which type of reaction does this equation represent

Neporo4naja [7]

this could be a substitution reaction. as you will locate, between the hydrogen's on the propane chain replaced into substituted for a Br from Br2. that's particularly no longer a addition reaction! addition reactions artwork once you have a AlkENE! by using fact that's an AlkANE it would not have a double bond to act as a nucleophile to attack the Br2 (which might act as a electrophile to boot reactions).

8 0

4 years ago