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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 clown in the park gave a child a balloon filled with 7.00 L of helium at 301 K. What will the volume of the balloon be when th

dybincka [34]

Answer:

6.9L

Explanation:

6 0

3 years ago

How do invertebrates affect the oxygen levels of compost and soil?.

Digiron [165]

Invertebrates affect the oxygen levels of compost and soil by reducing it.

<h3>How respiration works?</h3>

Cellular respiration is the process by which living organisms obtain energy by breaking down food molecules in their cells.

Every living organism undergoes the process of cellular respiration and make use of oxygen molecules in the process.

This means that invertebrates will make use of the oxygen gas present in compost and the soil for respiration, hereby reducing it.

Learn more about respiration at: brainly.com/question/480297

8 0

2 years ago

If the atmospheric pressure on the top of a mountain is 0.2396 atm, what is the pressure in Kilopascal

arsen [322]

Answer:

24.28 kPa

Step-by-step explanation:

You want to convert atmospheres to kilopascals.

The conversion factor is 101.325 kPa = 1 atm

p = 0.2396 atm × (101.325 kPa/1 atm)

p = 24.28 kPa

4 0

3 years ago

Read 2 more answers

consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of

MakcuM [25]

Answer:

CaO is the limiting reagent

Theoritical yield = 25.71 g

% Yield = 75.44%

Explanation:

1 mole = Molar mass of the substance

Molar Mass of CaO = 56 g/mol

Molar Mass of CaCO3 = 100 g/mol

Molar mass of CO2 = 44 g/mol

The balanced Equation is :

$CaO + CO_{2}\rightarrow CaCO_{3}$

1 mole of CaO reacts with = 1 mole of CO2

56 g of CaO reacts with = 44 g of CO2

1 g of CaO reacts with =

$\frac{44}{56}$

= 0.785 g of CO2

So,

14.4 g of CaO must react with = (14.4 x 0.785) g of CO2

= 11.31 g of CO2

Needed = 11.31 g

Available CO2 = 13.8 g (given)

So CO2 is in excess , hence CaO is the limiting reagent and product will produce from 14.4 g of CaO

$CaO + CO_{2}\rightarrow CaCO_{3}$

1 mole of CaO will produce 1 mole pf CaCO3

56 g of CaO produce = 100 g of CaCO3

1 g of CaO produce =

$\frac{100}{56}$

= 1.785 g of CaCO3

14.4 g of CaO will produce = (1.785 x 14.4) g of CaCO3

= 25.71 g of CaCO3

Theoritical Yield of CaCO3 = 25.71 g

Actual yield = 19.4 g

Percent Yield =

$\frac{Actual\ yield}{Theoritical\ yield}\times 100$

$\frac{19.4}{25.71}\times 100$

= 75.44 %

6 0

3 years ago

What is the empirical formula of a compound that has a pseudoformula of C3.5H8?

KIM [24]

Answer:

The ratio of carbon and hydrogen atoms = 3.5 : 8

= 7 : 16

Then,the empirical formula is C7H16

5 0

3 years ago