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stellarik [79]
3 years ago
6

Identify the term that matches each electrochemistry definition. The electrode where oxidation occurs Cathode The electrode wher

e reduction occurs Choose... An electrochemical cell powered by a spontaneous redox reaction Choose... An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction Choose... A chemical equation showing either oxidation or reduction Choose...
Chemistry
1 answer:
Luda [366]3 years ago
6 0

Answer:

An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction

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Why do we sometimes use roman numerals in naming ionic compounds?
Blababa [14]
Roman numerals are used in naming ionic compounds when the metal cation forms more than one ion. The metals that form more than one ion are the transition metals, although not all of them do this.
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3 years ago
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Which of the following is not a type of fault?
bekas [8.4K]

normal is the answer


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3 years ago
Find the final product for: Li + Br and show work.
miskamm [114]

Lithium is atomic number 3, so it has valency 1

While, Bromine is atomic number 35, and has valency 1

Lithium has an extra electron while Bromine need an electron, since they both need and have one electron, the form

LiBr (Lithium Bromide) where Li is +ve charged and Br is -ve charged

Happy to help :)

7 0
3 years ago
In the laboratory, a volume of 100 mL of sulfuric acid (H2SO4) is recorded. How many g are there of the liquid if its density is
ser-zykov [4K]

Answer:

\large \boxed{\text{183 g}}  

Explanation:

\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\1.83 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{100 cm}^{3}}\\\\m & = & \text{183 g}\\\end{array}\\\text{There are $\large \boxed{\textbf{183 g}}$ of sulfuric acid.}

8 0
3 years ago
Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s
NeX [460]

Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

3 0
3 years ago
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