A sample contains 430 mg of mercury. How many significant figures are in the number?

ANSWER Soon!!

vichka [17]

Answer:

1. 0.1683 mol

2. 1.191 g

3. 0.02695 mol

4. Na₂Cl₃

5. The empirical formula obtained is not correct. This is likely due to experimental errors, since much precision was required (the masses are determined in thousandths of grams).

Explanation:

1. How many moles of elemental sodium were used in the reaction?

Since all of the solid sodium is used up by the reaction, you can cancluate the number of moles of elemental sodium used dividing the mass by the molar mass:

number of moles = mass in grams / atomic mass

mass in grams = 0.3870 g (given)

atomic mass = 22.990 g/mol

number of moles = 0.3870 g / 22.990 g/mol = 0.1683 mol

2. What is the mass of chlorine gas used in the reaction?

a) Mass of chlorine gas introduced in the flask = mass of the stoppered flask after filling it with chlorine gas - mass of the empty flask with the sopper

Mass of chlorine gas introduced = 158.1743g - 156.1870g = 1.9873 g

b) Mass of chlorine gas unreacted = 0.7962 g (given)

c) Mass of chlorine gas used = mass of chorine gas introduced in the flask - mass of chlorine gass un reacted

Mass of chlorine gas used = 1.9873g - 0.7962g = 1.1911g

3. How many moles of chlorine were used in the reaction?

molar mass of chlorine gas, Cl₂ = 2 × 35.453 g/mol = 70.906 g/mol

number of moles = mass in grams / molar mass = 1.911g / 70.906g/mol = 0.02695 mol

4. What is the empirical formula of sodium chloride based on the experimental data?

Divide the number of moles of each element by the smalles number of moles:

Na: 0.01683 / 0.01683 = 1

Cl = 0.02695 / 0.01683 = 1.6

Multiply by 2 to obtain whole numbers:

Na = 2

Cl = 3.2 = 3

Empirical formula Na₂Cl₃

5. Was the empirical formula you obtained correct using the chemists data correct? Why?

No, the empirical formula you obtained using the chemists data is not correct, because the correct empirical formula of sodium chloride is NaCl.

That is, there is 1 atom of sodium per every atom of chlorine in one chemical formula of NaCl, but that is not reflected by the empirical formula Na₂Cl₃.

That is a demostration of big experimental errors. You can speculate that the errors are likely due to problems of procedure collecting the gas or errors in measuring the masses.

As you see, the masses are measured to thousandths of grams, which requires much precision; thus smalls absolute errors could produce huge relative errors.

7 0

2 years ago

For scuba dives below 150 ft, helium is often used to replace nitrogen in the scuba tank. If 15.2 g of He(g) and 30.6 g of O2(g)

abruzzese [7]

Answer:

see explanation below

Explanation:

To do this exercise, we need to use the following expression:

P = nRT/V

This is the equation for an ideal gas. so, we have the temperature of 22 °C, R is the gas constant which is 0.082 L atm / mol K, V is the volume in this case, 5 L, and n is the moles, which we do not have, but we can calculate it.

For the case of the oxygen (AW = 16 g/mol):

n = 30.6 / 32 = 0.956 moles

For the case of helium (AW = 4 g/mol)_

n = 15.2 / 4 = 3.8 moles

Now that we have the moles, let's calculate the pressures:

P1 = 0.956 * 0.082 * 295 / 5

P1 = 4.63 atm

P2 = 3.8 * 0.082 * 295 / 5

P2 = 18.38 atm

Finally the total pressure:

Pt = 4.63 + 18.38

Pt = 23.01 atm

7 0

2 years ago

Calculate the mass in grams of 0.800 mole of H2CO3. g

PilotLPTM [1.2K]

MH₂CO₃: (1g×2) + 12g + (16g×3) = 62 g/mol

1 mol --- 62g
0,8 mol -- X
X = 0,8×62
X = 49,6g

6 0

2 years ago

Read 2 more answers

Calculate the wavelength in meters of electromagnetic radiation that has a frequency of 1.09 × 10⁸ s⁻¹. (c = 3.00 X 10⁸ m/s)

fenix001 [56]

Answer:

2.75X10^16

Explanation:

C=FXwavelength

wavelength=C/F

7 0

2 years ago

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

a)

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d)

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

2 years ago