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2 years ago

When dilute HCL is added to zinc pieces taken in a test tube

Chemistry

1 answer:

Romashka-Z-Leto [24]2 years ago

6 0

Answer:

zince chloride will be formed and hydrogen gas will be librated.

Explanation:

When dilute HCl is added to zinc pieces, a rection will takes place as follows :

$Zn+HCl\rightarrow ZnCl_2+H_2$

It means that zinc chloride will form when zinc reacts with dilute HCL. Also hydrogen gas will produced.

As zinc is more reactive than hydrogen, it displaces hydrogen from its solution and forms zinc chloride. The form product is white in color and H₂ is an odorless gas.

Hence, zince chloride will be formed and hydrogen gas will be librated.

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1.Chlorobenzene, C6H5Cl, is used in the production of chemicals such as aspirin and dyes. One way that chlorobenzene is prepared

miv72 [106K]

Answer:

$m_{C_6H_5Cl}=65.7gC_6H_5Cl$

Explanation:

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In this case, according to the given balanced chemical reaction:

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$m_{C_6H_5Cl}=45.6gC_6H_6*\frac{1molC_6H_6}{78.11gC_6H_6 }*\frac{1molC_6H_5Cl}{1molC_6H_6}*\frac{112.55gC_6H_5Cl}{1molC_6H_5Cl} \\\\m_{C_6H_5Cl}=65.7gC_6H_5Cl$

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3 years ago

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Sholpan [36]

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atomic modifier

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3 years ago

Read 2 more answers

Calculate the theoretical yield of ammonia produced by the reaction of 100g of H2 gas and 200g of N2 gas

ipn [44]

To get the theoretical yield of ammonia NH3:
first, we should have the balanced equation of the reaction:
3H2(g) + N2(g) → 2NH3(g)
Second, we start to convert mass to moles
moles of N2 = N2 mass / N2 molar mass
= 200 / 28 = 7.14 moles
third, we start to compare the molar ratio from the balanced equation between N2 & NH3 we will find that N2: NH3 = 1:2 so when we use every mole of N2 we will get 2 times of that mole of NH3 so,
moles of NH3 = 7.14 * 2 = 14.28 moles
finally, we convert the moles of NH3 to mass again to get the mass of ammonia:
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3 years ago

A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an

erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

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When dilute HCL is added to zinc pieces taken in a test tube​

When dilute HCL is added to zinc pieces taken in a test tube