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4 years ago

The following reaction does not proceed to a product: H2O + Cu --> no reaction. Why is that?

2 answers:

5 0

I think the correct answer would be because copper has a lower activity than hydrogen and cannot replace the bonds in it. Substances that are not oxidizing do not react with copper since the redox potentials are very low. Hope this answers the question.

morpeh [17]4 years ago

5 0

The reason why H₂O +Cu does not proceed to a product is that

copper has a lower activity than hydrogen and cannot replace it (answer A)

Explanation

Replacement reaction is type of a reaction in which element react with compound and take place of another element in that compound.

That is the more reactive element replaces the less reactive element from it's compound.

Since H₂ is more reactive than Cu in reactivity series therefore no reaction occur between H₂O +Cu since Cu cannot replace H₂ from its compound.

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What is mean by kinetic energy

Vinvika [58]

The energy that an object has is moving

5 0

3 years ago

What volume of a 1.5 M KOH solution is needed to provide 3.0 moles of KOH?

Brums [2.3K]

Answer:

Volume = 2.0 liter of 1.5 M solution of KOH

Explanation:

1) Data:

a) Solution: KOH

b) M = 1.5 M

c) n = 3.0 mol

d) V = ?

2) Formula:

Molarity is a unit of concentration, defined as number of moles of solute per liter of solution:

M = n / V in liter

3) Calculations:

Solve for n: M = n / V ⇒ V = n / M

Substitute values: V = 3.0 mol / 1.5 M = 2.0 liter

You must use 2 significant figures in your answer: 2.0 liter.

8 0

3 years ago

Classify the following materials as solid, liquid, or gas at room temperature: milk, helium, granite, oxygen, steel, and gasolin

Sati [7]

Im not sure but milk is liquid, helium is gas, granite is solid, oxygen is gas, steel is solid, and gasiloine is liquid.

3 0

3 years ago

A 5.0 L sample of gas at 300. K is heated to 600. K. What will the new volume of the gas be?

Ainat [17]

Answer:

$V_2=10L$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the required new volume by using the Charles' law as a directly proportional relationship between temperature and volume:

$\frac{V_2}{T_2} =\frac{V_1}{T_1}$

In such a way, we solve for V2 and plug in V1, T1 and T2 to obtain:

$V_2=\frac{V_1T_2}{T_1}\\\\V_2=\frac{5.0L*600K}{300K}\\\\V_2=10L$

Regards!

4 0

3 years ago

It is proposed to use Liquid Petroleum Gas (LPG) to fuel spark-ignition engines. A typical sample of the fuel on a volume basis

Norma-Jean [14]

Answer:

The overall balanced combustion reaction is written as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2$

$(F/A)_{stoichiometric} = 0.0424$

$(A/F)_{stoichiometric} = 23.562$

the higher heating values $(HHV)_f$ per unit mass of LPG = 49.9876 MJ/kg

the lower heating values $(LHV)_f$ per unit mass of LPG = 46.4933 MJ/kg

Explanation:

The stoichiometric equation can be expressed as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> aCO_2 \ + \ bH_2O \ + \ cN_2$

Now, equating the coefficient of carbon; we have:

(0.7×3)+(0.05×4)+(0.25×3) = a

a = 3.05

Also, Equating the coefficient of hydrogen : we have:

(0.7 × 8) +(0.05 × 10) + ( 0.25 × 6) = 2 b

2b = 7.6

b = 3.8

Equating the coefficient of oxygen

2x = 2a + b

$x = \frac{2a+b}{2} \\ \\ x = \frac{2(3.05)+3.8}{2} \\ \\ x = 4.95$

Equating the coefficient of Nitrogen

$c = 3.76x \\ \\ c = 3.76 *4.95 \\ \\ c = 18.612$

Therefore, The overall balanced combustion reaction can now be written as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2$

Now; To determine the stoichiometric F/A and A/F ratios; we have:

$(F/A)_{stoichiometric} = \frac{n_f}{n_a } \\ \\ (F/A)_{stoichiometric} = \frac{1}{4.95*(1+3.76)} \\ \\ (F/A)_{stoichiometric} = 0.0424$

$(A/F)_{stoichiometric} = \frac{n_a}{n_f } \\ \\ (A/F)_{stoichiometric} = \frac{4.95*(1+3.76)}{1} \\ \\ (A/F)_{stoichiometric} = 23.562$

What are the higher and lower heating values per unit mass of LPG?

Let calculate the molecular mass of the fuel in order to determine their mass fraction of the fuel components.

Molecular mass of the fuel $M_f = (0.7*M_{C_3H_5} ) + (0.05 *M_{C_4H_{10}}) + (0.25*M _{C_3H_6})$

= 30.8 + 2.9 + 10.5

= 44.2 kg/mol

Mass fraction of the fuel components can now be calculated as :

$m_{C_3H_8} = \frac{30.8}{44.2} \\ \\ m_{C_3H_8} = 0.7 \\ \\ \\ m_{C_4H_{10}} = \frac[2.9}{44.2} \\ \\ m_{C_4H_{10}} = 0.06 \\ \\ \\ m_{C_3H_6} = \frac{10.5}{44.2} \\ \\ m_{C_3H_6} = 0.24$

Finally; calculating the higher heating values $(HHV)_f$ per unit mass of LPG; we have:

$(HHV)_f=(0.7 * HHV_{C_3H_8}) + (0.06 *HHV_{C_4H_{10}})+(0.24*HHV_{C_3H_6} \\ \\ (HHV)_f=(0.7*50.38)+(0.06*49.56)+(0.24*48.95) \\ \\ (HHV)_f=49.9876 \ MJ/kg$

calculating the lower heating values $(LHV)_f$ per unit mass of LPG; we have:

$(LHV)_f = (HHV)_f - \delta H_w \\ \\ (LHV)_f = (HHV)_f - [\frac{m_w}{m_f}h_{vap}] \\ \\ (LHV)_f = 49.9876 \ MJ/kg - [\frac{3.8*18}{44.2}*2.258 \ MJ/kg] \\ \\ (LHV)_f = 46.4933 \ M/kg$

7 0

4 years ago