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frutty [35]
2 years ago
13

The elements carbon and sulfur are solids at room temperature. Can you reasonably predict that a compound of these two elements

will also be a solid at room temperature?
Chemistry
1 answer:
tester [92]2 years ago
7 0
Yes I think your write
You might be interested in
Why don't solids change shape?
navik [9.2K]
Solids are compounds whose atomic bonds are rigid, which doesn't let the atoms move around freely
7 0
2 years ago
Read 2 more answers
What can be said about a reaction with H = 62.4 kJ/mol and S = 0.145 kJ/(mol·K)?
Zigmanuir [339]
Answer:

At 430.34 K the reaction will be at equilibrium, at  T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

Explanation:

1) Variables:

G = Gibbs energy
H = enthalpy
S = entropy

2) Formula (definition)

G = H + TS

=> ΔG = ΔH - TΔS

3) conditions

ΔG < 0 => spontaneous reaction
ΔG = 0 => equilibrium
ΔG > 0 non espontaneous reaction

4) Assuming the data given correspond to ΔH and ΔS

ΔG = ΔH - T ΔS = 62.4 kJ/mol + T 0.145 kJ / mol * K

=>  T = [ΔH - ΔG] / ΔS

ΔG = 0 =>  T = [ 62.4 kJ/mol - 0 ] / 0.145 kJ/mol*K = 430.34K

This is, at 430.34 K the reaction will be at equilibrium, at  T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

3 0
3 years ago
Read 2 more answers
What is the concentration in molarity of a solution that has 54.21g of calcium hydroxide dissolved in 560mL of water?
ra1l [238]
We know that Molarity = # mol/L

In the question, we are given grams - so we need to calculate the number of moles that we are dealing with. We can do this by first finding out what the molecular formula of calcium hydroxide is: Ca(OH)_{2

This means that we have 1 calcium and 2 hydroxides.

We can find the number of moles by taking the grams given divided by the molar mass of the molecule. We can find the molar mass of the molecule by taking each atomic mass of each atom (found on the periodic table) and adding them together:

Ca = 40.08g
O = 15.999g
H = 1.008g

We have 1 calcium, 2 oxygens, and 2 hydrogens. So let's add the atomic masses together:

40.08 + 2(15.999) + 2(1.008) = 74.094g

Now that we have the molar mass, we can find the number of moles due to already knowing the given amount in the question:

54.21g/74.09g = 0.73 moles of Ca(OH)_{2}

Now that we know the number of moles, we can solve for the molarity; The other part of the question gives us 560 mL of water - but this can quickly be converted to liters by moving the decimal to the left 3 places => 0.560 L

Knowing all of this information, we can plug it into the molarity equation:

Molarity = 0.73 mol/0.56 L
Molarity = 1.30

The molarity of the solution is 1.3 M.
4 0
2 years ago
(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW
Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = <em>2.3x10⁻³M  InF₃. </em>Thus:

[In⁺³] = 2.3x10⁻³M  InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M  In⁺³

[F⁻] = 2.3x10⁻³M  InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M  In⁺³] × [7.0x10⁻³M F⁻]³ = <em>7.9x10⁻¹⁰</em>

<em></em>

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, <em>solubility is 6.31x10⁻⁶M</em>

3 0
3 years ago
Q1:What is one result of meiosis?
anzhelika [568]

Answer:

1.  the end result of meiosis is haploid daughter cells with chromosomal combinations different from those originally present in the parent.

2.  Prophase, metaphase, and telophase

5 0
3 years ago
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