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2 years ago

What does a low number on the pH scale say about a solution?

Chemistry

2 answers:

irina [24]2 years ago

8 0

Answer:D The solution is an acid

Explanation:The lower the pH of a substance the stronger the acidic level.

vlabodo [156]2 years ago

6 0

Answer:

d. the solution is a acid

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What is the difference between atomic mass, relative atomic mass and average atomic mass

vesna_86 [32]

$\bold{\huge{\underline{\purple{ Answer }}}}$

Difference between Atomic mass, relative atomic mass and average atomic mass :-

<h3>Atomic Mass :-</h3>

Atomic mass is the mass of neutrons and protons present in the nucleus of an atom .
It is always calculated for a single element and having direct value
For isotopes also, the atomic mass is calculated separately . Example :- Carbon 12 , carbon 13 and carbon 14 have different atomic mass. 
The SI unit of Atomic mass is " u" and "amu"

<h3>Relative Atomic mass :-</h3>

Relative atomic mass is mean mass of the atoms of an element which is compared to the 1/12th mass of carbon - 12 .
Carbon - 12 is taken as a relative when we calculate the relative atomic mass of any element
For calculating relative atomic mass, we need to know the masses, percentage and abundance of all types of elements
Relative atomic mass is a dimension less quantity

<h3>Average Atomic Mass :-</h3>

Average atomic mass is the average mass of an atoms of a particular element by considering it's isotopes
While we calculate average atomic mass is a standardized number. Whereas, Average atomic mass sometimes varies geologically .
It also includes percentage, abundance and masses of given element .
In average atomic mass, We do not compare mean value with the 1/12 mass of carbon - 12
The unit of Average atomic mass is "Amu" or " u " .

5 0

2 years ago

29. A gas has a volume of 1.75 L at -23°C and 150.0 kPa.

arsen [322]

The answer for the following mention bellow.

Therefore the final temperature of the gas is 260 k

Explanation:

Given:

Initial pressure ( $P_{1}$ ) = 150.0 kPa

Final pressure ( $P_{2}$ ) = 210.0 kPa

Initial volume ( $V_{1}$ ) = 1.75 L

Final volume ( $V_{2}$ ) = 1.30 L

Initial temperature ( $T_{1}$ ) = -23°C = 250 k

To find:

Final temperature ( $T_{2}$ )

We know;

According to the ideal gas equation;

P × V = n × R ×T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

We know;

$\frac{P*V}{T}$ = constant

$\frac{P_{1} }{P_{2} }$ × $\frac{V_{1} }{V_{2} }$ = $\frac{T_{1} }{T_{2} }$

Where;

( $P_{1}$ ) represents the initial pressure of the gas

( $P_{2}$ ) represents the final pressure of the gas

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

( $T_{1}$ ) represents the initial temperature of the gas

( $T_{2}$ ) represents the final temperature of the gas

So;

$\frac{150 * 1.75}{210 * 1.30}$ = $\frac{260}{T_{2} }$

( $T_{2}$ ) =260 k

Therefore the final temperature of the gas is 260 k

3 0

2 years ago

How long does it take to electroplate 0.5 mm of gold on an object with a surface area of 31 cm^^ from an Au3+(aq) solution with

konstantin123 [22]

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = $a=31 cm^2$

Thickness of the gold plating = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

$V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3$

Density of the gold = $d=19.3 g/cm^3$

$m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g$

Moles of gold = $\frac{29.915 g}{197 g/mol}=0.152 mol$

$Au^{3+}+3e^-\rightarrow Au$

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

$\frac{3}{1}\times 0.152 mol=0.456 mol$ of electrons

Number of electrons = N = $0.456\times \times 6.022\times 10^{23}$

Charge on single electron = $q=1.6\times 10^{-19} C$

Total charge required = Q

$Q=N\times q$

Amount of current passes = I = 8 Ampere

Duration of time = T

$I=\frac{Q}{T}$

$T=\frac{N\times q}{I}$

$=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s$

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

7 0

2 years ago

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of

dsp73

The first thing we need to do here is to recognize the unit of molarity and the units of the given percentage of nitric acid.

Molarity is mol HNO3 / L of solution. This is our aim

The given percentage is 0.68 g HNO3/ g solution

multiplying this with density to convert g solution into mL solution and dividing with the molecular weight of HNO3 (63 g/mol) to convert g HNO3 to mol. Therefore we obtain

0.016 mol/ mL or 16.23 mol/ L (M)

6 0

3 years ago

Which contains 3 significant figures.

jolli1 [7]

B, E, F have 3 significant figures

4 0

3 years ago