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abruzzese [7]
3 years ago
9

Please help with all 3 parts!

Chemistry
2 answers:
Murljashka [212]3 years ago
8 0

Answer:

1:Part A.

\bold{42.2 g C_{12}H_{22}O_{11} \:in \:528 g H₂O}

Mass Percent=\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}

=\frac{42.2}{528}*100\%=\bold{\underline{7.99\: or \:8\%}}

Part B.

\bold{198\:m g\: C_{6}H_{12}O_{6} \:in\:4.71 g\: H₂O}

mass of solute: 198mg

mass of solvent :4.71g=4710g

Mass Percent=\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}

=\frac{198}{4710}*100\%=\bold{\underline{4.20\%}}

Part C.

\bold{8.85 g NaCl \:in \:190 g\: H₂O}

Mass Percent=\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}

=\frac{8.85}{190}*100\%=\bold{\underline{4.66\%}}

Alecsey [184]3 years ago
6 0

Answer:

It will help you !!!!!!!!!!

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How many different types of elements are there in acetone C3H6O?
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En un experimento hacemos reaccionar 12 g de carbono con 32 g de oxígeno para formar dióxido de carbono. Razona si podemos saber
max2010maxim [7]

Answer:

La masa de óxido de carbono iv formado es 44 g.

Explanation:

En esta pregunta, se nos pide calcular la masa de óxido de carbono iv formado a partir de la reacción de masas dadas de carbono y oxígeno.

En primer lugar, necesitamos escribir una ecuación química equilibrada.

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De la ecuación, 1 mol de carbono reaccionó con 1 mol de oxígeno para dar 1 mol de óxido de carbono iv.

Ahora, si marca las masas en la pregunta, verá que corresponde a la masa atómica y la masa molar de la molécula de carbono y oxígeno, respectivamente. ¿Qué indica esto?

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5 0
3 years ago
In the following overall chemical reaction, aluminum displaces chromium from chromium(III) oxide and forms aluminum oxide.
mafiozo [28]

Answer:

Change in enthalpy for the reaction is -536 kJ

Explanation:

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  • Take reaction (1a) and divide stoichiometric coefficients by 2
  • Take reverse reaction (2a) and divide stoichiometric coefficient by 2
  • Then add these two modified elementary steps to get overall chemical reaction
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2Al+\frac{3}{2}O_{2}\rightarrow Al_{2}O_{3}.....\Delta H_{1}=\frac{-3351.4}{2}kJ=-1675.7kJ

Cr_{2}O_{3}\rightarrow 2Cr+\frac{3}{2}O_{2}....\Delta H_{2}=\frac{2279.4}{2}kJ=1139.7kJ

--------------------------------------------------------------------------------------------------------------

2Al+Cr_{2}O_{3}\rightarrow 2Cr+Al_{2}O_{3}....\Delta H=\Delta H_{1}+\Delta H_{2}=-1675.7+1139.7kJ=-536kJ

6 0
3 years ago
How do you know anything?
r-ruslan [8.4K]
By studying and learning by using your brain to accomplish new goals. :)
(if that makes sense)
Hope this helps :3
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