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babunello [35]
3 years ago
10

Standard reduction potentials are based on which element?

Chemistry
2 answers:
KonstantinChe [14]3 years ago
6 0
They're based on hydrogen.
Nadusha1986 [10]3 years ago
3 0

Answer:

Hydrogen

Explanation:

Standard reduction potential is a measure of the tendency of an atom to attract electrons and get reduced. Higher the reduction potential i.e. more positive the value greater will be the tendency to get reduced. These values are measured at 298 K and 1 atm and expressed in terms of volts.

The standard reduction potential of Hydrogen is taken as zero and the values of all other elements are deduced based on hydrogen. These are tabulated in the thermochemical reduction potential table. The reduction equation for hydrogen is:

H2(g) + 2e-  → 2H+           where E⁰ = 0.00 Volts

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Weather, erosion and deposition will change the environment over time. When water runs over rocks it can smooth out jagged edges
Pachacha [2.7K]

Answer:

erosion.

Explanation:

7 0
3 years ago
if 98% of atoms of li have a mass of 7 and 2% have a mass of 6 what roughly will the relative atomic mass be​
Nastasia [14]

Answer: 6.98

The Relative Atomic Mass of Li = (98*7 + 2*6)/98+2

RAM of Li = (686+12)/100

RAM of Li= 6.98

6 0
3 years ago
If the methane contained in 2.50 L of a saturated solution at 25 ∘C was extracted and placed under STP conditions, what volume w
Sunny_sXe [5.5K]

116.6 x 10^{ -3} is the volume of methane contained in 2.50 L of a saturated solution at 25 ∘C that was extracted and placed under STP conditions.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

The solubility of methane in water at 25 degrees Celcius is 1.3 x 10^{ -3} M. It implies   1.3 x 10^{ -3} moles of methane are dissolved in one litre of water.

The number of moles of methane in 4 L of water can be calculated as follows:

Moles of methane = \frac{1.3 . 10^{ -3} }{1L} x\frac{4L}{1L}

Moles of methane = 5.2 x 10^{ -3}

STP refers to standard temperature and pressure. Under STP conditions, the temperature of the substance is  0 degrees celcius and its pressure is 1 atm.

An ideal gas is an imaginary gas comprising of a large number of randomly moving particles and the motion between such articles is considered to be perfectly elastic. The ideal gas equation describes the relationship between pressure, volume, temperature and number of moles of a gas.

The expression for ideal gas equation is as follows:

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 1 atm

V= ?

R= 0.082057338 \;L \;atm \;K^{-1}mol^{-1}

T=273K

n=?

Putting value in the given equation:

\frac{PV}{RT}=n

5.2 x 10^{ -3} = \frac{1 \;atm\; X \;V}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 273}

V= 116.6 x 10^{ -3}

Hence, 116.6 x 10^{ -3} is the volume of methane contained in 2.50 L of a saturated solution at 25 ∘C was extracted and placed under STP conditions.

Learn more about the ideal gas here:

brainly.com/question/27691721

#SPJ1

4 0
2 years ago
A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in a spheric
Luba_88 [7]

Complete Question

A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in a spherical air tank that measures 74.0 wide. The biologist estimates she will need 2600 L of air for the dive. Calculate the pressure to which this volume of air must be compressed in order to fit into the air tank. Write your answer in atmospheres. Round your answer to significant digits.

Answer:

The pressure required is P_2= 12.2 \ atm

Explanation:

Generally the volume of a sphere is mathematically denoted as

             V_s = \frac{4}{3} * \pi r^3

Substituting r =  \frac{d}{2} = \frac{74}{2} = 37cm

          V_s = \frac{4}{3} * 3.42 * (37)^2

               V_s = 2.121746 *10^5 cm^3

Converting to Liters

               V_s = \frac{2.121746 *10^5}{1000}

                V_s= 212.1746L

Assume that the pressure at which the air is given to the diver is 1 atm when the air was occupying a volume of 2600L

So

From Charles law

               P_1V_1 = P_2 V_s

Substituting  V_1 =2600 L ,   P_1 = 1 atm , V_s  =212.1746L , and making P_2 the subject we have

                 P_2 = \frac{P_1 * V_1}{V_s}

                     = \frac{1 * 2600}{212.1746}

                    P_2= 12.2 atm

               

6 0
3 years ago
Find the molality of this aqueous solution 15.0% by mass kBr (119g/mol).
Vsevolod [243]

Answer:

we will take a 100g sample of this solution for our convenience

so , there is 15 g kBr in this 100g solution

we know that molality is the number if moles of solute / mass of solvent in kg

we need to find the number of moles in 15g kBr

no of moles = 15/119 s

moles  = 0.126 moles/ 100g

multiplying both the numerator and the denominator by 10 to get 1 kg in denominator

=  1.26 moles / 1 kg

Hence, the molality is 1.26

would appreciate a brainliest

3 0
4 years ago
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