Only Extensive properties of piece will be changed. i.e Mass, Volume, Length etc.
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: <span>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. </span><span>Os Exemplos de Cálculos de Massa molar: </span> NaCl <span>, </span> o Ca (OH) 2 <span>, </span> K4 [Fe (CN) 6] <span>, </span> CuSO4 * 5H2O <span>, </span> água <span>, </span> ácido nítrico <span>, </span> permanganato de potássio <span>, </span> etanol <span>, </span> 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. <span>Os Exemplos de Cálculos de peso molecular:</span> C [14] o [16] 2 <span>,</span> S [34] O [16] 2 <span>.</span> Definição de massa molecular, o peso molecular, uma massa molar e do peso molar <span><span><span>Massa molecular</span></span></span><span><span> ( </span></span><span><span><span>peso molecular</span></span></span><span><span> ) e uma massa de uma molécula de uma substancia e e Expressa nsa unificadas unidades de massa atómica (u). </span></span><span><span>(1 u E igual a 1/12 da massa de hum átomo de carbono-12) </span></span><span><span><span>Massa Molar</span></span></span><span><span> ( </span></span><span><span><span>molar peso</span></span></span><span><span> ) E uma massa de Uma toupeira de Uma substancia ê ê expresso em g / mol.</span></span><span> Pesos dos Átomos e isótopos São de </span><span>NIST Artigo </span><span>.</span> DEIXE Seu comentário <span>Sobre a SUA Experiência com uma calculadora de Peso Molecular </span><span>Pesos </span><span>moleculares de Aminoacidos: Relacionados</span>
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Answer:
The water is completely vaporized at this stage.
Explanation:
The complete question is
If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is
-boiling
-completely vaporized
-frozen solid
-decomposed
-still a liquid
Energy added = 50 kJ = 50000 J
mass of water = 15.5 g = 0.0155 kg
temperature of water = 10 °C
We know that the energy posses by a mass of water at a given temperature is given as
H = mcT
where H is the energy possessed by the mass of water
m is the mass of the water
c is the specific heat capacity of water = 4200 J/ kg- °C
T is the temperature of the water
substituting values, the energy of this amount of water is
H = 0.0155 x 4200 x 10 = 651 J
If 50 kJ is added to the water, the energy increases to
50000 J + 651 J = 50651 J
Temperature of this water at this stage will be gotten from
H = mcT
we solve for the new temperature
50651 = 0.0155 x 4200 x T
50651 = 65.1 x T
T = 50651/65.1 = 778.05 °C
This temperature is well over 100 °C, which is the vaporization temperature of water, but less than 3000 °C for its molecules to decompose.
Answer:
= 67.79 g
Explanation:
The equation for the reaction is;
4Cr(s)+3O2(g)→2Cr2O3(s)
The mass of O2 is 21.4 g, therefore, we find the number of moles of O2;
moles O2 = 21.4 g / 32 g/mol
=0.669 moles
Using mole ratio, we get the moles of Cr2O3;
moles Cr2O3 = 0.669 x 2/3
=0.446 moles
but molar mass of Cr2O3 is 151.99 g/mol
Hence,
The mass Cr2O3 = 0.446 mol x 151.99 g/mol
<u> = 67.79 g
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