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Goshia [24]
3 years ago
6

1. What is the density of a 20 gram of 40 ml liquid?

Chemistry
1 answer:
____ [38]3 years ago
4 0

Answer:

1. 0.5 g/mL

2. 0.88 g/cm^3

3. It has the greatest mass to volume ratio

4. Incomplete

5. H^+ (aq)+OH^-(aq)\rightleftharpoons H_2O (l)

6. 1

7. Gloves, goggles, coat, mask

8. Flush with tap water for at least 15 minutes

9. The facts are listed below

10. Mass and volume

12. All matter consists of moving particles, the degree of their movement is directly proportional to their kinetic energy

Explanation:

1. In order to solve for density, we should know that density is the ratio between mass and volume of a liquid. In this case, we're given both measures: given mass of m = 20 g and volume of V = 40 mL, we may simply apply the equation of density described here:

d=\frac{m}{V}

Substituting the variables, we obtain:

d=\frac{m}{V}=\frac{20 g}{40 mL}=0.5 g/mL

2. Given a mixture of several liquids, it's important to understand that liquids with a greater density will tend to form a bottom layer of a solution, while liquids with a lower density will tend to form a top layer of a solution. Here we have a liquid with a density of d_1 = 1.0 g/cm^3 and another liquid with a density of d_2 = 0.88 g/cm^3. Notice that d_1 > d_2.

This implies that the liquid with a density of 0.88 g/cm^3 would be on top, as its density is lower than the density of the other liquid with a density of 1.0 g/cm^3.

3. The solid phase is not always, but typically denser than liquids or gases. There are some exceptions to this rule, for example, ice, a solid phase of water, is less dense than liquid water.

However, for the majority of cases this statement is true. Remember that solid phases are the most ordered phases with atoms being packed closely to each other. In liquids, atoms are more dispersed with distances between them being greater than those in solids. Similarly, gases have the greatest distances between gas atoms among all three phases.

Since density is directly proportional to mass, let's say we take the same volume of a solid, a liquid and a gas. For the same volume, since we'll have a greater number of solid atoms than for a liquid or a gas (because solid atoms are more closely packed with lower average distances between the atoms), the mass to volume ratio will be the greatest for solids.

4. This seems to be an incomplete question.

5. In order to balance the following ionic equation, we need to follow the mass and charge balancing rules. Firstly, expand a water molecule showing the individual parts of it:

H-OH

Secondly, notice that we need to add a hydroxide anion to the proton, so that we obtain the same number of protons and hydroxide anions on the left side, as well as the number of hydrogen and oxygen atoms on the right. This way, the net charge on the left hand side (0) and the net charge on the right hand side (0) are equal, so the charge is balanced as well. We obtain:

H^+ (aq)+OH^-(aq)\rightleftharpoons H_2O (l)

6. We should be familiar with the ionization constant of water in the context of this problem. It is defined as the product between the hydronium ions and hydroxide ions and is a constant number at some given temperature. For pure water, the concentration of hydronium ions is balanced by the concentration of hydroxide anions to yield a neutral pH value, meaning the ratio of one with respect to the other would be 1.

For example, at room temperature, the ionization constant of water is defined as:

K_w=[H^+][OH^-]=10^{-14}

Since we have pure water:

[H^+]=[OH^-]=\sqrt{10^{-14}}=10^{-7}

Then the ratio is:

[tex]\frac{[H^+]}{[OH^-]}=\frac{10^{-7}}{10^{-7}}=1

7 to 12. The questions are explained in the file attached.

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A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr
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Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

Concentration = Moles / volume

[NO] = 0.10 mol / 1L = 0.10 M

[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

[NO] at the equilibrium is 0.062 M

This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

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