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2 years ago

Calculate the density of a material that has mass of 52.457g and a volume of 13.5 cm3

Chemistry

1 answer:

Nesterboy [21]2 years ago

5 0

1 cm3 = 1 ml
Density is g/ml

So this is 52.457g / 13.5ml = 3.89 g/ml

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A 45 mL sample of nitrogen gas is cooled from 135ºC to 15C in a container that can contract or expand at constant pressure. Wha

Vanyuwa [196]

Answer:

V₂ =31.8 mL

Explanation:

Given data:

Initial volume of gas = 45 mL

Initial temperature = 135°C (135+273 =408 K)

Final temperature = 15°C (15+273 =288 K)

Final volume of gas = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 45 mL × 288 K / 408 k

V₂ = 12960 mL.K / 408 K

V₂ =31.8 mL

8 0

2 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

3 years ago

What is the mass of a sample of a sample of iron (cp=0.44j/g•°C) when it is heated from 60°to 160°C. The heat is transferred is

scZoUnD [109]

Q=mcat

6600=(m)(.44)(100)

6600=(44m)

150= m

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3 years ago

21.

Bond [772]

The answer is A. The pH decreases

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2 years ago

How would it have been if particles of matter had no space?

iragen [17]

Answer:

Matter is made of small particles of atoms or molecules. ... The particles in the solid are touching with very little space between them. The particles in a liquid usually are still touching but there are some spaces between them. The gas particles have big distances between them.

6 0

2 years ago