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3 years ago

Which of the following is an empirical formula? OP205 O C2H602 ON3F6 OC8H18

Chemistry

1 answer:

jarptica [38.1K]3 years ago

7 0

Answer:

None are empirical formulas

Explanation:

All are actual compounds. An example of an empirical formula could be CH2O, the empirical formula for carbohydrates like glucose (C6H12O6).

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If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + NaI → PbI2 + NaNO3

ss7ja [257]

Answer:- 27.7 grams of $PbI_2$ are produced.

Solution:- The balanced equation is:

$Pb(NO_3)_2+2NaI\rightarrow PbI_2+2NaNO_3$

let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.

Molar mass of $Pb(NO_3)_2 = 207.2+2(14.01)+6(16) = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2$ = 207.2+2(126.90) = 461 gram per mol

let's do the calculations for the grams of the product for the given grams of each of the reactant:

$28.0gPb(NO_3)_2(\frac{1molPb(NO_3)_2}{331.22gPb(NO_3)_2})(\frac{1molPbI_2}{1molPb(NO_3)_2})(\frac{461gPbI_2}{1molPbI_2})$

= $39.0gPbI_2$

$18.0gNaI(\frac{1molNaI}{149.89gNaI})(\frac{1molPbI_2}{2molNaI})(\frac{461gPbI_2}{1molPbI_2})$

= $27.7gPbI_2$

From above calculations, NaI gives least amount of $PbI_2$ , so the answer is, 27.7 g of $PbI_2$ are produced.

8 0

3 years ago

A 2.50-l volume of hydrogen measured at â196 Â°c is warmed to 100 Â°c. calculate the volume of the gas at the higher temperature

ivann1987 [24]

To solve this we assume that the hydrogen gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

T1 / V1 = T2 / V2

V2 = T2 x V1 / T1

V2 = (100 + 273.15) K x 2.50 L / (-196 + 273.15) K

V2 = 12.09 L

Therefore, the volume would increase to 12.09 L as the temperature is increased to 100 degrees Celsius.

5 0

3 years ago

Myth: A dead organism is the same as a nonliving thing in science. Fact: Evidence:

Ilya [14]

I recently did this assignment!

Instructions: Read each myth (untruth). Reword it to make a factual statement. Then, give two to three reasons why the myth is untrue. Use complete sentences and support your answer with evidence, using your own words.

________________________________________

Answer:

Myth: A dead organism is the same as a nonliving thing in science.

o Fact: In science, dead is the same as nonliving.

o Evidence: Things that are nonliving never had the characteristics of life, and never will. Things that are dead once did have the characteristics of life, but when they die, they lose some of the characteristics. That is why dead and non-living are NOT the same thing.

Hope this helped!

Have an Amazing day!

~Lola

4 0

3 years ago

Read 2 more answers

Calculate the formula for the following hydrate composed of 76.9% CaSO3 and 23.1%H2O

artcher [175]

Assume there is 100g of the substance at first

5 0

3 years ago

Read 2 more answers

Under standard-state conditions, which of the following species is the best reducing agent? a. Ag+ b. Pb c. H2 d. Ag e. Mg2+

eimsori [14]

Answer: The correct answer is Option b.

Explanation:

Reducing agents are defined as the agents which help the other substance to get reduced and itself gets oxidized. They undergo oxidation reaction.

$X\rightarrow X^{n+}+ne^-$

For determination of reducing agents, we will look at the oxidation potentials of the substance. Oxidation potentials can be determined by reversing the standard reduction potentials.

For the given options:

Option a: $Ag^+$

This ion cannot be further oxidized because +1 is the most stable oxidation state of silver.

Option b: $Pb$

This metal can easily get oxidized to $Pb^{2+}$ ion and the standard oxidation potential for this is 0.13 V

$Pb\rightarrow Pb^{2+}+2e^-;E^o_{(Pb/Pb^{2+})}=+0.13V$

Option c: $H_2$

This metal can easily get oxidized to $H^{+}$ ion and the standard oxidation potential for this is 0.0 V

$H_2\rightarrow 2H^++2e^-;E^o_{(H_2/H^{+})}=0.0V$

Option d: $Ag$

This metal can easily get oxidized to $Ag^{+}$ ion and the standard oxidation potential for this is -0.80 V

$Ag\rightarrow Ag^{+}+e^-;E^o_{(Ag/Ag^{+})}=-0.80V$

Option e: $Mg^{2+}$

This ion cannot be further oxidized because +2 is the most stable oxidation state of magnesium.

By looking at the standard oxidation potential of the substances, the substance having highest positive $E^o$ potential will always get oxidized and will undergo oxidation reaction. Thus, considered as strong reducing agent.

From the above values, the correct answer is Option b.

8 0

3 years ago