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mario62 [17]
3 years ago
10

2.0 grams of ZnSO4 reacts completely with Li2CO3; how many grams of Li2SO4 would be produced?

Chemistry
1 answer:
grin007 [14]3 years ago
3 0
Moles of ZnSO4 = Mass/Mr
= 2/(65.4 + 32.1 + (16 x 4))
= 0.012383.... mol

Since there is no chemical equation provided, I will assume that the ratio of ZnSO4:Li2CO3 is 1:1

Therefore there are 0.012383 mil of Li2SO4 as well
So the mass would be Moles x Mr = 0.012383.... x ((6.9 x 2) +32.1 + (4x16))
= 1.360 g of Li2SO4 produced
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I will give brainliest. If you burn the carbon in limited air, the reaction is
Fynjy0 [20]

This reaction is different in that the carbon undergoes an incomplete combustion as opposed to complete combustion where carbon is fully oxidized. A caveat: incomplete combustion products in general can be difficult to predict without sufficient information, as it's not uncommon to obtain a mixture of different products.

Here, we are told that solid carbon is burned in limited air to produce a gas. I am presuming that, in the equation that's given, the "0" represents a blank where you must fill in a chemical symbol. In this case, our equation would be: 2C(s) + O₂(g) → 2CO(g).

There is not enough information here to provide the numerical answers to the two questions. From the words in the question (e.g., "is different" and "this time"), it would seem that this question is an excerpt from a larger or preceding question where specific numbers had been provided or computed.

However, it's possible to make some general observations on how one may go about answering these questions <em>if </em>one had more information.

Since we're to assume that oxygen is the limiting reagent, if one is given the amount of solid carbon (either in mass, moles, or number of atoms), it's possible to determine the moles of CO(g) that's produced since C and CO have an equal stoichiometric ratio. So, for example, if one burns 2 moles of C(s), then 2 moles of CO(g) would be produced.

<em><u>But</u></em>, there is still not enough information to compute the volume of CO gas if this is the line of questioning. We don't know, for instance, the temperature or pressure of the reaction conditions. In fact, the only way it would be possible to answer this would be if you were given beforehand a conversion factor that relates the volume of CO(g) to its quantity (e.g., to assume that one mole of gas occupies <em>x </em>liters).

As for the second question, this would depend on what you know about the quantity of the C(s) reacted and/or the quantity (or volume, from question a) of CO(g) produced. If you can get the number of moles of C(s) reacted or CO(g) produced, the number of moles of O₂(g) used up: It would be half the number of moles of C(s) reacted or half the number of moles of CO(g) produced). <u>Again</u>, it's impossible to determine the volume of O₂(g) using just the information provided here, so I suspect that you must have further information relating gas quantity to volume. As we did with CO(g), the volume of O₂(g) used up can be found using whatever conversion factor you have.

If you have any further information or questions, please feel free to follow up.  

6 0
2 years ago
4. A 14.5-L balloon is in the air where it is 20.0 C at 0.980-atm. A wind passes over and the balloon's pressure becomes 740.0-m
jeka94

The temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.

<h3>What is Combined gas law?</h3>

Combined gas law put together both Boyle's Law, Charles's Law, and Gay-Lussac's Law. It states that "the ratio of the product of volume and pressure and the absolute temperature of a gas is equal to a constant.

It is expressed as;

P₁V₁/T₁ = P₂V₂/T₂

Given the data in the question;

  • Initial volume V₁ = 14.5L
  • Initial pressure P₁ = 0.980atm
  • Initial temperature T₁ = 20.0°C = 293.15K
  • Final volume V₂ = 14.3L
  • Final pressure P₂ = 740.mmHg = 0.973684atm
  • Final temperature T₂ = ?

We substitute our given values into the expression above.

P₁V₁/T₁ = P₂V₂/T₂

( 0.980atm × 14.5L )/293.15K = ( 0.973684atm × 14.3L )/T₂

14.21Latm / 293.15K = 13.92368Latm / T₂

14.21Latm × T₂ = 13.92368Latm × 293.15K

14.21Latm × T₂ = 4081.72679LatmK

T₂ = 4081.72679LatmK / 14.21Latm

T₂ = 287.24K

T₂ = 14.09°C

Therefore, the temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.

Learn more about the combined gas law here: brainly.com/question/25944795

#SPJ1

4 0
2 years ago
A 0.652-g sample of a pure strontium halide reacts with excess sulfuric acid. the solid strontium sulfate formed is separated, d
seraphim [82]

Answer:

The formula of the original halide is SrCl₂.

Explanation:

  • The balanced equation of this reaction is:

SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X is the halide.

  • From the equation stichiometry, 1.0 mole of strontium halide will result in 1.0 mole of SrSO₄.
  • The number of moles of SrSO₄ <em>(n = mass/molar mass) </em>= (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
  • The number of moles of SrX are  4.11 x 10⁻³ moles from the stichiometry of the balanced equation.
  • n = mass / molar mass, n =  4.11 x 10⁻³ moles and mass = 0.652 g.
  • The molar mass of SrX₂ = mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
  • The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
  • The atomic mass of halide X = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2  g/mole = 35.5 g/mole.
  • This is the atomic mass of Cl.
  • <em>So, the formula of the original halide is SrCl₂</em>.
4 0
3 years ago
A 50.0 mL graduated cylinder has a mass of 65.1 g. When it is filled with an unknown liquid to the 49.3 mL mark, the cylinder an
Delvig [45]

<u>Answer:</u> The density of liquid is 1.12g/cm^3

<u>Explanation:</u>

We are given:

Mass of cylinder, m_1 = 65.1 g

Mass of liquid and cylinder combined, M = 120.5 g

Mass of liquid, m_2 = (M-m_1)=(120.5-65.1)g=55.4g

To calculate density of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

We are given:

Mass of liquid = 55.4 g

Volume of liquid = 49.3 mL = 49.3 cm^3    (Conversion factor:  1mL=1cm^3 )

Putting values in above equation, we get:

\text{Density of liquid}=\frac{55.4g}{49.3cm^3}\\\\\text{Density of liquid}=1.12g/cm^3

Hence, the density of liquid is 1.12g/cm^3

3 0
3 years ago
The entropy of an exothermic reaction decreases. This reaction will be spontaneous under which of the following temperatures?
kakasveta [241]

Answer:Low temperatures

Explanation:

∆G= ∆H-T∆S

If ∆H is negative (exothermic reaction), then in order to maintain ∆G<0 which is the condition for spontaneity; T must decrease. This is because, decrease in T will keep the difference of ∆H and T∆S at a negative value in order to satisfy the above stated condition for spontaneity.

3 0
3 years ago
Read 2 more answers
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