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3 years ago

How many moles are in 5.03 x 1023 molecules?

Chemistry

1 answer:

AlladinOne [14]3 years ago

4 0

Answer:

Some formulas for calculating mole are

Mole = Mass/ Molar mass

Mole = no of particles / avogadros constant

NB : no of particles can be no of atoms , no of ions , or no of molecules 2. Avogadros number or constant = 6.02 times 10 ^23

so we will be using the second formula

Mole = no of particles / avogadros constant

Mole = 5.03 x 10 ^23/6.02 x10^23

Mole = 8.355x10^45

hope it helps :)

Explanation:

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Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2

PIT_PIT [208]

<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ$

Hence, the standard heat for the given reaction is -138.82 kJ

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3 years ago

What number of atoms of phosphorus are present in 1.00g of each of the compounds in exercise 40?

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Answer:

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Explanation:

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2 years ago

What is the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure? As air

Marina86 [1]

We need to know the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure.

The relationship is: As air pressure in an area increases, the density of the gas particles in that area increases.

For any gaseous substance, density of gas is directly proportional to pressure of gas.

This can be explained from idial gas edquation:

PV=nRT

PV= $\frac{w}{M}$ RT [where, w= mass of substance, M=molar mass of substance]

PM= $\frac{w}{V}$ RT

PM=dRT [where, d=density of thesubstance]

So, for a particular gaseous substance (whose molar mass is known), at particular temperature, pressure is directly related to density of gaseous substance.

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3 years ago

Determine if the data are qualitative or quantitative.

jonny [76]

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3 years ago

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What is the quantity of heat (in kJ) associated with cooling 185.5 g of water from 25.60°C to ice at -10.70°C?Heat Capacity of S

Cerrena [4.2K]

Taking into account the definition of calorimetry, sensible heat and latent heat, the amount of heat required is 37.88 kJ.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Sensible heat</h3>

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

<h3>Latent heat</h3>

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

<u><em>25.60 °C to 0 °C</em></u>

First of all, you should know that the freezing point of water is 0°C. That is, at 0°C, water freezes and turns into ice.

So, you must lower the temperature from 25.60°C (in liquid state) to 0°C, in order to supply heat without changing state (sensible heat).

The amount of heat a body receives or transmits is determined by:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

c= Heat Capacity of Liquid= 4.184 $\frac{J}{gC}$
m= 185.5 g
ΔT= Tfinal - Tinitial= 0 °C - 25.60 °C= - 25.6 °C

Replacing:

Q1= 4.184 $\frac{J}{gC}$ × 185.5 g× (- 25.6 °C)

Solving:

<u><em>Change of state</em></u>

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

In this case, you know:

n= 185.5 grams× $\frac{1mol}{18 grams}$ = 10.30 moles, where 18 $\frac{g}{mol}$ is the molar mass of water, that is, the amount of mass that a substance contains in one mole.