Answer:
The volume of the gas will be 78.31 L at 1.7 °C.
Explanation:
We can find the temperature of the gas by the ideal gas law equation:
Where:
n: is the number of moles
V: is the volume
T: is the temperature
R: is the gas constant = 0.082 L*atm/(K*mol)
From the initial we can find the number of moles:
Now, we can find the temperature with the final conditions:
The temperature in Celsius is:
Therefore, the volume of the gas will be 78.31 L at 1.7 °C.
I hope it helps you!
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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Here is the complete question.
Benzalkonium Chloride Solution ------------> 250ml
Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.
Sig: Dilute 10ml to a liter and apply to affected area twice daily
How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?
(A) 1700 mL
(B) 29.4 mL
(C) 17 mL
(D) 294 mL
Answer:
(B) 29.4 mL
Explanation:
1 L = 1000 mL
1:200 solution implies the 
in 200 mL solution.
200 mL of solution = 1g of Benzalkonium chloride
1000 mL will be 
200mL × 1g = 1000 mL × x(g)
x(g) = 
x(g) = 0.2 g
That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.
∴ 
y(g) = 
y(g) = 5g of benzalkonium chloride.
Now, at 17% concentrate contains 17g/100ml:
∴ the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;
= 
z(mL) = 
z(mL) = 29.41176 mL
≅ 29.4 mL
Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride
“Bonding molecular orbitals are formed by... in-phase combinations of atomic wave functions, and electrons in these orbitals stabilize a molecule.”
Answer:
positively charged elctrons
