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Olin [163]
2 years ago
12

7. A solution containing 90grams of KNO3 per 100. grams of H2O at

Chemistry
1 answer:
Tomtit [17]2 years ago
5 0

Answer:

(4) concentrated and supersaturated

Explanation:

At 50.°C, 90g of KNO3 lies above the solubility curve [on the Regents Reference Table G]. This indicates that the solution is supersaturated, meaning it contains more solute than will naturally dissolve, and was formed when a saturated solution cooled. Furthermore, the percent concentration of this solution is 90% KNO3 making this solution concentrated. This can be calculated using the formula for mass percent concentration.

Percent Mass = <u>Mass of Solute (g)</u> x 100

                         Mass of Solution (g)

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C3H8(g) + 5O2(g) ⟶ 3CO2(g) + 4H2O(g) H = -2220 kJ If 865.9 g of H2O is produced during this combustion, how much heat is generat
dem82 [27]

Answer:

3 × 10⁴ kJ

Explanation:

Step 1: Write the balanced thermochemical equation

C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(g) ΔH = -2220 kJ

Step 2: Calculate the moles corresponding to 865.9 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

865.9 g × 1 mol/18.02 g = 48.05 mol

Step 3: Calculate the heat produced when 48.05 moles of H₂O are produced

According to the thermochemical equation, 2220 kJ of heat are evolved when 4 moles of H₂O are produced.

48.05 mol × 2220 kJ/4 mol = 2.667 × 10⁴ kJ ≈ 3 × 10⁴ kJ

5 0
2 years ago
Please help !!!
stiv31 [10]

Answer:

2KClO3 》》2KCl +3O2

C+ O2》》CO2

number of C moles

Required O2 moles (According to the mole ratio )

Relevant to the first equation, find the moles the KClO3, which is used to produce that amount of O2 moles

Now you can find the mass of KClO3

I mentioned the useful steps which can guide you to get the answer.

Explanation:

5 0
2 years ago
For each chemical equation below, write the number of product molecules that will form from the reaction.
Gekata [30.6K]

Answer:

a. 5 moles of CO₂

b. 2 moles of NaCl

c. 3 moles of H₂CO₃

Explanation:

1a. Carbon reacts with oxygen to produce carbon dioxide according to equation below:

C + O₂ ----> CO₂

From the equation of reaction, one mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide.

In the given reaction, therefore, 5 moles of carbon will only react with 5 moles of oxygen to produce 5 moles of carbon dioxide. Oxygen is in excess, carbon is the limiting reactant and 5 moles of carbon dioxide are produced. The equation is given below :

5 C + 6 O₂ ---> 5 CO₂

1b. Sodium reacts with chlorine to produce sodium chloride according to the equation below:

2 Na + Cl₂ ---> NaCl

From the equation of reaction, 2 moles of sodium reacts with one mole of chlorine to produce one mole of sodium chloride.

In the given reaction, therefore, 4 moles of sodium will react with only 2 moles of chlorine to produce 2 moles of sodium chloride. Chlorine is in excess, sodium is the limiting reactant and 2 moles of sodium chloride are produced. The equation is given as follows: 4 Na + 8 Cl₂ ---> 2 NaCl

1c. Carbon dioxide reacts with water to produce carbonic acid according to the equation: CO₂ + H₂O ---> H₂CO₃

From the equation of reaction one mole of carbon dioxide reacts with one mole of water to produce one mole of carbonic acid.

In the given reaction, therefore, 3 moles of carbon dioxide will react with only 3 moles of water to produce 3 moles of carbonic acid. Water is in excess, carbon dioxide is the limiting reactant and 3 moles of carbonic acid are produced. The equation is given as follows: 3 CO₂ + 4 H₂O ---> 3 H₂CO₃

4 0
3 years ago
Why do you think we are able to study the atom in the modern world?
andriy [413]
Advancement in technology to help with detailed studies
6 0
2 years ago
2. Which pair of elements does not form an ionic bond?
Gekata [30.6K]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the concept of Bond Formation.

Answer is :-

B.) O and Cl

These both are Non-Metals with high Electronegativity nature, hence none of them can share electrons to foram any bond to eventually form a Electro valentines Compound.

5 0
3 years ago
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