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3 years ago

Which phrase describes radiation?

Chemistry

1 answer:

vesna_86 [32]3 years ago

8 0

Answer:

the process in which energy is emitted by one object, transmitted through space, and absorbed by another

Explanation:

Radiation is energy that comes from a source and travels through space and may be able to penetrate various materials. Light, radio, and microwaves are

types of radiation that are called nonionizing.

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What is a way of moving electricity along a path called? *

OLEGan [10]

The answer is circuit I believe.

4 0

3 years ago

Starting with 250 mL of a 0.250 M solution of HBr; a) Calculate the initial pH of the solution.b) Calculate the pH after adding

maxonik [38]

Answer:

pH = 0.602

pH = 1.5

pH = 7

pH = 12.1

Explanation:

a ) To calculate the pH, use the following equation:

pH = -log [H+]

Hbr is a strong acid, so the [H+] concentration can be calculated as follow:

[HBr] = 0.250 M

As acid Hbr:

Hbr = H+ + Br-

As strong acid HBr dissociates at all, so

[HBr] = [H+] = 0.250 M

So the pH:

pH = -log [0.250 M] = 0.602

b) Calculate the pH after adding 250 mL of 0.125M NaOH

In this point, the reactions starts:

HBr + NaOH = H2O + NaBr

- First, we gonna find the mol of each reactant:

HBr:

mol = [M] × L

mol = 0.250 M × 0.250 L

mol = 0.0625 mol HBr

NaOH:

mol = [M] × L

mol = 0.125 M × 0.250L

mol = 0.03125 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.0625 mol HBr – 0.03125 mol HBr = 0.03125 mol HBr

These are the moles free in the solution, and we going to use them to calculate the pH:

pH = -log [ H]

pH = - log [ 0.03125 ] = 1.5

c) Calculate the pH after adding 500 mL of 0.125M NaOH

NaOH:

mol = [M] × L

mol = 0.125 M × 0.500L

mol = 0.0625 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.0625 mol HBr – 0.0625 mol HBr = 0 HBr

pH = 7

d) Calculate the pH after adding 600 mL of 0.125M NaOH.

NaOH:

mol = [M] × L

mol = 0.125 M × 0.600L

mol = 0.075 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.075 mol NaOH – 0.0625 mol NaOH = 00125 NaOH

These are the moles free in the solution, and we going to use them to calculate the pH:

pOH = -log [ OH]

pOH = - log [ 0.0125 ] = 1.90

pH + pOH = 14

pH = 14- pOH = 14 – 1.90 = 12.1

4 0

3 years ago

True or false. an object's weight is constant in all circumstances

Sindrei [870]

Answer: False

Explanation: False, what if its in space?

3 0

2 years ago

This is a mixture that has the same composition throughout.

FinnZ [79.3K]

Answer: Homogeneous mixture

Explanation:

Homogeneous mixtures are those mixtures in which the dispersed phase is uniformly distributed throughout the dispersion medium. The dispersed phase has uniform composition throughout the dispersion medium and thus there is no physical boundary between the dispersed phase and the dispersion medium. Example: salt in water

Heterogeneous mixtures are those mixtures in which the dispersed phase is not uniformly distributed throughout the dispersion medium. The dispersed phase does not has uniform composition throughout the dispersion medium and thus there is a distinct physical boundary between the dispersed phase and the dispersion medium. Example: Sand in water

5 0

3 years ago

What is the voltage of the voltaic cell Zn|Zn2+||Cu2+|Cu at 298 K if [Zn2+] = 0.2 M and [Cu2+] = 4.0 M? Cu2+ + 2e- → Cu Eo = +0.

OLga [1]

Answer : The voltage of the voltaic cell is 1.14 V

Explanation :

From the given cell representation, we conclude that

The copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$