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2 years ago

Which compound will give positive Tollen's test?

1 answer:

5 0

Tollen's test is an organic test that determines whether the sample is an adehyde or not. When aldehydes are present, the reaction forms a silver compound which refers to the other name of Tollen's as silver mirror test. The aldehyde among the choices is C. pentanal

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Consider the reaction: P4 + 6Cl2 = 4PCl3.

likoan [24]

Answer:

The answer to your question is below

Explanation:

Consider the reaction: P4 + 6Cl2 = 4PCl3.

a. How many grams of Cl2 are needed to react with 20.00 g of P4? ___68.7 g___________

P4 + 6Cl2 = 4PCl3

4(31) ---------- 12(35.5)

20 ---------- x

x = 20(12x35.5) / 4(31)

x = 8520 / 124

x = 68.7 g

b. You have 15.00 g. of P4 and 22.00 g. of Cl2, identify the limiting reactant and calculate the grams of PCl3 that can be produced as well as the grams of excess reactant remaining. LR____________ grams PCl3 _________ grams excess reactant ___________

P4 + 6Cl2 = 4PCl3

124g 426 g 4(31 + 3(35.5)) = 550g

15g 22g

I will use P4 to find the limiting reactant

x = (15 x 426) / 124 = 51.5 The limiting reactant is Chlorine

because we need 51.5 g and we only have 22g

Excess reactant

x = (22 x 124) / 426 = 6.4 g of P4

Excess P4 = 15 g - 6.4 = 8.6 g of P4 in excess

Grams of PCl3 produced

426 g of Cl2 ---------------- 550 g of PCl3

22g of Cl2 ------------- - x

x = (22 x 550) / 426 = 28.4 g of PCl3

c. If the actual amount of PCl3 recovered is 16.25 g., what is the percent yield? ______________

% yield = (16.25 - 28.4) / 28.4 x 100

% yield = 42.8

d. Given 28.00 g. of P4 and 106.30 g. of Cl2, identify the limiting reactant and calculate how many grams of the excess reactant will remain after the reaction. LR ______________ grams excess reactant

Limiting reactant

124g of P4 ------------- 426 g 6Cl2

28g --------------- x

x = (28 x 426) / 124

x = 96.2 g of Cl2 and we have 106.3 so Chlorine is the excess reactant and P4 is the limiting reactant.

Excess reactant = 106.3 - 96.2 = 10.1 g of Cl2 in excess

4 0

2 years ago

A student carries out this chemical reaction in a beaker:

nydimaria [60]

Answer:The correct answer is option (C).

Explanation:

$Zn+CuCl_2\rightarrow Cu+ZnCl_2$

After the reaction between Zinc and copper(II) chloride, Zinc chloride and copper will be obtained as a product. This reaction occurs because zinc is more reactive than copper which displaces the copper from its aqueous solution.

The student will expect to find copper metal at bottom of the beaker with aqueous solution of zinc chloride.

Hence the correct answer is option (C).

7 0

2 years ago

A metal X from two oxide A and B .3.oogm of A and B contain 0.72 and 1.16g of oxygen respectively.calculate the maases of metal

devlian [24]

Answer:

Explanation:

Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.

Metal X can form 2 oxides (A and B).

A + B = 3g

The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.

The mass of metal X in the two oxides will be the same because it's the same metal.

Thus, we represent the mass of the metal in the two oxides as 2X.

2X + 0.72 + 1.16 = 3

2X + 1.88 = 3

2X = 3 - 1.88

2X = 1.12

X = 0.56

<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>

Thus, mass of metal (X) in 1g of oxygen in A is

0.56g ⇒ 0.72g

X ⇒ 1

X = 1 × 0.56/0.72

X = 0.78 g

Hence, 0.78g of the metal will combine with 1g of oxygen for A

Also, mass of metal (X) in 1g of oxygen in B is

0.56g ⇒ 1.16g

X ⇒ 1g

X = 1×0.56/1.16

X = 0.48 g

Thus, 0.48g of the metal will combine with 1g of oxygen for B

6 0

2 years ago

Part B Write the balanced chemical equation for the neutralization reaction that occurs when an aqueous solution of hydrobromic

Doss [256]

Answer:

HBr(aq) + LiOH(aq) → LiBr(aq) + H2O(l)

Explanation:

A neutralization reaction is a process in which an acid, aqeous HBr reacts completely with an appropriate amount of base, aqueous LiOH to produce salt, aqueous LiBr and water, liquid H2O only.

HBr(aq) + LiOH(aq) → LiBr(aq) + H2O(l)

Acid + base → Salt + Water.

During this reaction, the hydrogen ion, H+, from the HBr is neutralized by the hydroxide ion, OH-, from the LiOH to form the water molecule, H2O.

Thus, it is called a neutralization reaction.

4 0

2 years ago

For the reaction 2kclo3(s)→2kcl(s)+3o2(g) calculate how many grams of oxygen form when each quantity of reactant completely reac

barxatty [35]

First, we need to get the molar mass of:

KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol

KCl =39.1 + 35.5 = 74.6 g/mol

O2 = 16*2 = 32 g/mol

From the given equation we can see that:

every 2 moles of KClO3 gives 3 moles of O2

when mass = moles * molar mass

∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g

and the mass of O2 then = 3 mol * 32g/mol = 96 g

so, 245.2 g of KClO3 gives 96 g of O2

A) 2.72 g of KClO3:

when 245.2 KClO3 gives → 96 g O2

2.72 g KClO3 gives → X

X = 2.72 g KClO3 * 96 g O2/245.2 KClO3

= 1.06 g of O2

B) 0.361 g KClO3:

when 245.2 g KClO3 gives → 96 g O2

0.361 g KClO3 gives → X

∴ X = 0.361g KClO3 * 96 g / 245.2 g

= 0.141 g of O2

C) 83.6 Kg KClO3:

when 245.2 g KClO3 gives → 96 g O2

83.6 Kg KClO3 gives → X

∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3

= 32.7 Kg of O2

D) 22.4 mg of KClO3:

when 245.2 g KClO3 gives → 96 g O2

22.4 mg KClO3 gives → X

∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3

= 8.8 mg of O2

7 0

3 years ago