What element is used in bright flashing advertising signs

The pOH of a solution is 6.0. Which statement is correct? Use p O H equals negative logarithm StartBracket upper O upper H super

Katen [24]

Answer:

The pH of the solution is 8.

Explanation:

To which options are correct, let us determine the concentration of the hydroxide ion, [OH-] and the pH of the solution. This is illustrated below:

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

pOH = –Log [OH-]

pOH = 6

6 = –Log [OH-]

–6 = Log [OH-]

[OH-] = Antilog (–6)

[OH-] = 1x10^–6 mol/L

2. The pH of the solution can be obtained as follow:

pH + pOH = 14

pOH = 6

pH + 6 = 14

pH = 14 – 6

pH = 8.

From the calculations made above,

[OH-] = 1x10^–6 mol/L

pH = 8.

Therefore, the correct answer is:

The pH of the solution is 8

3 0

1 year ago

A sample of an alloy of aluminum contains 0.0898 mol Al and 0.0381 mol Mg. What are the mass percentages of Al and Mg in the all

blondinia [14]

Answer:

Al 72.61%

Mg 27.39%

Explanation:

To obtain the mass percentages, we need to place the individual masses over the total mass and multiply by 100%.

If we observe clearly, we can see that the parameters given are the moles. We need to convert the moles to mass.

To do this ,we need to multiply the moles by the atomic masses. The atomic mass of aluminum is 27 while that of magnesium is 24.

Now, the mass of aluminum is thus = 27 * 0.0898 = 2.4246g

The mass of magnesium is 0.0381 * 24 = 0.9144g

We can now calculate the mass percentage.

The total mass is 0.9144 + 2.4246 = 3.339g

% mass of Al = 2.4246/3.339 * 100 = 72.61%

% mass of Mg = 0.9144/3.39 * 100 = 27.39%

7 0

2 years ago

Read 2 more answers

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit

Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

$x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L$

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0

2 years ago

A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water

lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

$Q_{water} -Q_{sample} = 0$

$Q_{water} = Q_{sample}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{sample}$ - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

$m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)$

Where:

$m_{s}$ , $m_{w}$ - Mass of water and the sample of aluminium, measured in kilograms.

$c_{p,s}$ , $c_{p,w}$ - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

$T_{s}$ , $T_{w}$ - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

$T$ - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

$(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}$

$T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}$

Given that $m_{s} = 0.380\,kg$ , $m_{w} = 2.40\,kg$ , $c_{p,s} = 910\,\frac{J}{kg\cdot K}$ , $c_{p,w} = 4186\,\frac{J}{kg\cdot K}$ , $T_{s} = 378\,K$ and $T_{w} = 273\,K$ , the final temperature of the system is:

$T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}$

$T = 276.494\,K$

The equilibrium temperature of the system is 276.494 Kelvin.

7 0

2 years ago

Solid aluminum metal and diatomic chlorine gas react spontaneously to form a solid product. Give the balanced chemical equation

ValentinkaMS [17]

When solid aluminum metal is reacted with diatomic chlorine gas, solid aluminum chloride is formed. This reaction is an example of synthesis or chemical combination in which two elements, aluminum and chlorine combine to form a new compound aluminum chloride.

Word equation: Aluminum (s)+ Chlorine (g)---> Aluminum chloride(s)

Molecular formula of the product formed is $AlCl_{3}$ .

Therefore the balanced chemical equation representing the reaction of solid aluminum with gaseous dichlorine can be represented as,

$2Al(s) + 3Cl_{2}(g)-->2AlCl_{3}(s)$

8 0

2 years ago