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Gnoma [55]
3 years ago
8

2. Arnie is going to dissolve 2.65g of zinc using 6.0M hydrochloric acid. A) what volume of the acid will he need to dissolve al

l the zinc? B) He wants to collect the gas produced by the reaction in a container, determine the volume of container he will need if the pressure is 628 mmHg and temperature is 23ᵒC. C) What is the mass of the zinc chloride he will produce? You will need a balanced equation and a gas law for this question.
Chemistry
1 answer:
fenix001 [56]3 years ago
5 0

Answer:

Explanation:

Zn + 2HCl = ZnCl₂ + H₂

A ) mole of Zn = 2.65 / 65

= .04 mol

mole of HCl required = .04 x 2 mol

.08 mol

If v be the volume required

v x 6 = .08

v = .0133 liter

= 13.3 cc

B )

volume of gas at NTP :

moles of gas  obtained = .04 moles

= 22.4 x .04 liter

= .896 liter

we have to find this volume at given temperature and pressure

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

\frac{760\times.896}{273} =\frac{628\times V_2}{296}

V_2 = 1.175 liter.

C )

.04 mole of zinc chloride will be produced

mol weight of zinc chloride

= 65 + 35.5 x 2

= 136 gm

.04 mole = 136 x .04

= 5.44 gm

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What mass of natural gas (CH4) must you burn to emit 264 kJ of heat?
mariarad [96]
The heat released by cpmolete the combustion of organic products with oxygen is called heat of combustion.

Here 1 mol of CH4 realesed 802.3 KJ

To emit 264 kJ you multiply you need (1 mol of CH4/802.3 kJ)* 264 kJ = 0.329 mol of CH4

The molar mass, MM, of CH4 is 12 g/mol + 4*1g/mol = 16 g/mol

The to obtain the mass multiply the number of moles times the molar mass:

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7 0
3 years ago
g A piece of solid Zn metal is put into an aqueous solution of Cu(NO3)2. Write the net ionic equation for any single-replacement
Nimfa-mama [501]

Answer:

Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)

Explanation:

Let's consider the molecular single displacement equation between Zn and Cu(NO₃)₂

Zn(s) + Cu(NO₃)₂(aq) ⇒ Zn(NO₃)₂(aq) + Cu(s)

The complete ionic equation includes all the ions and insoluble species.

Zn(s) + Cu²⁺(aq) + 2 NO₃⁻(aq) ⇒ Zn²⁺(aq) + 2 NO₃⁻(aq) + Cu(s)

The net ionic equation includes only the ions that participate in the reaction and insoluble species.

Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)

4 0
3 years ago
The hydroboration of an alkene occurs in ___________ which places the boron of the borane on the ___________ carbon of the doubl
Nana76 [90]

Answer: The hydroboration of an alkene occurs in TWO CONCERTED STEP which places the boron of the borane on the LESS SUBSTITUTED carbon of the double bond. The oxidizing agent then acts as a nucleophile, attacking the electrophilic BORON and resulting in the placement of a hydroxyl group on the attached carbon. Thus, the major product of the hydroboration oxidation reaction DOES NOT follow Markovnikov's rule.

Explanation:

Hydroboration is defined as the process which allows boron to attain the octet structure. This involves a two steps pathway which leads to the production of alcohol.

--> The first step: this involves the initiation of the addittion of borane to the alkene and this proceeds as a concerted reaction because bond breaking and bond formation occurs at the same time.

--> The second step: this involves the addition of boron which DOES NOT follow Markovnikov's rule( that is, Anti Markovnikov addition of Boron). This is so because the boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon.

Note: The Markovnikov rule in organic chemistry states that in alkene addition reactions, the electron-rich component of the reagent adds to the carbon atom with fewer hydrogen atoms bonded to it, while the electron-deficient component adds to the carbon atom with more hydrogen atoms bonded to it.

8 0
3 years ago
The first two steps in the industrial synthesis of nitric acid produce nitrogen dioxide from ammonia: 2N0(g) +02(g) 2NO2 (g) The
Tresset [83]

The question is incomplete, complete question is :

The first two steps in the industrial synthesis of nitric acid produce nitrogen dioxide from ammonia:

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Step 2 :  2NO(g) +O_2(g) \rightleftharpoons 2NO_2 (g)

The net reaction is:

4NH_3(g)+7O_2(g)\rightleftharpoons 4NO_2(g)+6H_2O(g)

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Answer:

Equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 \& K_2:

K=K_1\time (K_2)^2

Explanation:

Step 1 : 4NH_3(g)+5O_2(g)\rightleftharpoons 4NO(g)+6H_2O(g)

Expression of an equilibrium constant can be written as:

K_1=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}

Step 2 :  2NO(g) +O_2(g) \rightleftharpoons 2NO_2 (g)

Expression of an equilibrium constant can be written as:

K_2=\frac{[NO_2]^2}{[NO]^2[O_2]}

The net reaction is:

4NH_3(g)+7O_2(g)\rightleftharpoons 4NO_2(g)+6H_2O(g)

Expression of an equilibrium constant can be written as:

K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}

Multiply and divide [NO]^4;

K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO]^4}{[NO]^4}

K=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO_2]^4}{[NO]^4}

K=K_1\times \frac{[NO_2]^4}{[O_2]^2[NO]^4}

K=K_1\times (\frac{[NO_2]^2}{[O_2]^1[NO]^2})^2

K=K_1\time (K_2)^2

So , the equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 \& K_2:

K=K_1\time (K_2)^2

4 0
3 years ago
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