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3 years ago

What is the formula mass for fe(no3)3

2 answers:

5 0

Calculando a massa molar (molar peso) Para Calcular a massa molar de hum Composto químico, ponha SUA fórmula E clique em 'Calcular'. Na Fórmula química that rápido Você PODE USAR: QUALQUÉR elemento químico Grupos Funcionais: D, Ph, Me, Et, Bu, AcAc, Para, Ts, Tos, Bz, TMS, tBu, Bzl, Bn, Dmg PARENTESIS () UO colchetes [] . Nomes Comuns de Compostos. Os Exemplos de Cálculos de Massa molar: NaCl , o Ca (OH) 2 , K4 [Fe (CN) 6] , CuSO4 * 5H2O , água , ácido nítrico , permanganato de potássio , etanol , frutose . Peso Computação molecular (massa molecular) para calcular o peso molecular de um composto químico entrar em sua fórmula, especifique seu número de massa isotópica depois de cada elemento entre colchetes. Os Exemplos de Cálculos de peso molecular: C [14] o [16] 2 , S [34] O [16] 2 . Definição de massa molecular, o peso molecular, uma massa molar e do peso molar Massa molecular ( peso molecular ) e uma massa de uma molécula de uma substancia e e Expressa nsa unificadas unidades de massa atómica (u). (1 u E igual a 1/12 da massa de hum átomo de carbono-12) Massa Molar ( molar peso ) E uma massa de Uma toupeira de Uma substancia ê ê expresso em g / mol. Pesos dos Átomos e isótopos São de NIST Artigo . DEIXE Seu comentário Sobre a SUA Experiência com uma calculadora de Peso Molecular Pesos moleculares de Aminoacidos: Relacionados

ahrayia [7]3 years ago

4 0

The formula mass of fe(no3)3 is179.87.

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A 50-mL beaker only has a scale that measures 10, 20, 30, and 40 mL. What is the uncertainty associated with the 50 mL beaker.

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3 years ago

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3 years ago

When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain

yKpoI14uk [10]

Answer:

For a: The number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

For b: The mass of nitric acid formed is 54.81 grams

For c: The mass of nitric acid formed is 206 grams

Explanation:

The given chemical reaction follows:

$3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)$

For a:

By Stoichiometry of the reaction:

1 mole of water reacts with 3 moles of nitrogen dioxide

So, 0.250 moles of water will react with $\frac{3}{1}\times 0.250=0.75mol$ of nitrogen dioxide

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.75 moles of nitrogen dioxide will contain $0.75\times 6.022\times 10^{23}=4.52\times 10^{23}$ number of molecules

Hence, the number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

For b:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of nitrogen dioxide = 60.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{60.0g}{46g/mol}=1.304mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 2 mole of nitric acid

So, 1.304 moles of nitrogen dioxide will produce = $\frac{2}{3}\times 1.304=0.870$ moles of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.870 moles

Putting values in equation 1, we get:

$0.870mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.870mol\times 63g/mol)=54.81g$

Hence, the mass of nitric acid formed is 54.81 grams

For c:

For nitrogen dioxide:

Given mass of nitrogen dioxide = 225 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{225g}{46g/mol}=4.90mol$

For water:

Given mass of water = 55.2 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{55.2g}{18g/mol}=3.06mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide reacts with 1 mole of water

So, 4.90 moles of nitrogen dioxide will react with = $\frac{1}{3}\times 4.90=1.63mol$ of water

As, given amount of water is more than the required amount. So, it is considered as an excess reagent.

Thus, nitrogen dioxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 mole of nitrogen dioxide produces 2 moles of nitric acid

So, 4.90 moles of nitrogen dioxide will produce $\frac{2}{3}\times 4.90=3.27mol$ of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 3.27 moles

Putting values in equation 1, we get:

$3.27mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(3.27mol\times 63g/mol)=206g$

Hence, the mass of nitric acid formed is 206 grams

7 0

3 years ago