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Elanso [62]
2 years ago
14

Which of the following are examples of physical properties of ethanol? Select all that apply.

Chemistry
1 answer:
bazaltina [42]2 years ago
7 0
Ethanol is: flammable, liquid at room temperature, the boiling point is 78.37 ° C.
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What is the position of an element in the periodic table if its electron configuration is 1sÆ 2sÆ 2pÒ 3sÆ 3p¢?
Maurinko [17]
Sodium I think I'm not sure
6 0
3 years ago
A solution has a [H3O+] of 1x 10-5 M. What is the [OH-] of the solution? (5 points)
White raven [17]

Answer:

1x10^–9 M

Explanation:

From the question given,

Concentration of hydronium ion, [H3O+] = 1x10^-5 M.

Concentration of Hydroxide ion, [OH-] =..?

The concentration of the hydroxide ion, [OH-] can be obtained as follow:

[H3O+] x [OH-] = 1x10^–14

1x10^-5 M x [OH-] = 1x10^–14

Divide both side by 1x10^-5

[OH-] = 1x10^–14 / 1x10^-5

[OH-] = 1x10^–9 M

8 0
3 years ago
A 44% wt/wt solution of H2SO4 has a density of 1.343g/ml. What mass of H2SO4 is 60ml of this solution?
horsena [70]

<u>Answer:</u> The mass of sulfuric acid present in 60 mL of solution is 34.1 grams

<u>Explanation:</u>

We are given:

44 % (m/m) solution of sulfuric acid. This means that 44 grams of sulfuric acid is present in 100 grams of solution.

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

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

Density of solution = 1.343 g/mL

Mass of solution = 100 g

Putting values in above equation, we get:

1.343g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.343g/mL}=77.46mL

To calculate the mass of sulfuric acid present in 60 mL of solution, we use unitary method:

In 77.46 mL of solution, mass of sulfuric acid present is 44 g

So, in 60 mL of solution, mass of sulfuric acid present will be = \frac{44g}{77.46mL}\times 60mL=34.1g

Hence, the mass of sulfuric acid present in 60 mL of solution is 34.1 grams

3 0
3 years ago
Which has the highest boiling point .33 m NH3 or .10 m Na2SO4​
Aleksandr-060686 [28]

Answer:

0.33 mol/kg NH₃

Explanation:

Data:

     b(NH₃) = 0.33 mol/kg

b(Na₂SO₄) = 0.10 mol/ kg

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor — the number of moles of particles you get from a solute.

(a) For NH₃,

The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.  

1 mol NH₃ ⟶  1 mol particles

i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water

(b) For Na₂SO₄,

Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)

1 mol Na₂SO₄ ⟶ 3 mol particles

i = 1 and ib = 3 × 0.10 = 0.30 mol particles per kilogram of water

The NH₃ has more moles of particles, so it has the higher boiling point.

5 0
3 years ago
Where do producers belong on the food chain?
Akimi4 [234]
The bottom of the food chain
4 0
2 years ago
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