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4 years ago

15

PRoblem 1: For which of the following activities might you want to hire a chemist?

1 answer:

Evgesh-ka [11]4 years ago

4 0

1) B, Not 100% sure though

2)Building a rocket is a physical change technically its not changing its composition its just changing its appearance, the fuel for it causes a chemical change, there is an answer that is more clear, Grinding chalk into a fine powder only changes its shape which is a physical change.

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59 Q.No. 28 b Hydrogen peroxide can act as oxidizing as well as reducing agent with example.

professor190 [17]

Answer:

Hydrogen peroxide can function as an oxidizing agent as well as reducing agent.

H2O2 act as oxidizing agent in acidic medium.

Explanation:

Example : 2FeSO4 +H2SO4 +H2O2 —>

(ferrous sulphate)

Fe2(SO4)3 +2H2O

(ferric sulphate)

3 0

2 years ago

Can anyone here help me with chemistry

Kay [80]

The answer is 20g N2

5 0

3 years ago

If you want to prepare 80.0 mL of 4.00M acid ,How many mL of 12.4 M HCl are required ?

Oksi-84 [34.3K]

$M_{A}V_{A}=M_{B}V_{B}\\(80.0)(4.00)=V_{B}(12.4)\\V_{B}=\frac{(80.0)(4.00)}{12.4} \approx \boxed{25.8 \text{ mL}}$

5 0

2 years ago

How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and

Kay [80]

Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

$(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)$

The expression used will be:

$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]$

where,

$Q$ = heat released for the reaction = ?

m = mass of benzene = 94.4 g

$c_{p,s}$ = specific heat of solid benzene = $1.51J/g^oC=1.51J/g.K$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC=1.73J/g.K$

$\Delta H_{fusion}$ = enthalpy change for fusion = $-9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g$

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

$Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]$

$Q=-29427.312J=-29.4kJ$

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

3 0

3 years ago

Can someone help me please :)

Fudgin [204]

I think it’s the third option but I’m not entirely sure

3 0

3 years ago

Read 2 more answers