All categories

2 years ago

Explain why the product at the negative electrode is not always a metal.

Chemistry

1 answer:

Elena-2011 [213]2 years ago

3 0

Answer:

Whether you get the metal or hydrogen during electrolysis depends on the position of the metal in the reactivity series: the metal will be produced if it is less reactive than hydrogen. hydrogen will be produced if the metal is more reactive than hydrogen.

You might be interested in

15. What volume of CCI, (d = 1.6 g/cc) contain

anastassius [24]

Answer:

$\boxed{\text{(3) 9.6 L}}$

Explanation:

1. Moles of CCl₄

$n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}$

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

$m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}$

4. Volume of CCl₄

$V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}$

4 0

2 years ago

Why does galium melt

levacccp [35]

Answer:

Like most other metals, Gallium is solid at room temperature (or liquid if it is too hot in your room). But, if it is held [in hands] for long enough, it melts in your hands, and doesn't poison you like Mercury would. This is because of its unusually low melting point of (~29 degree Centigrade).

- It melts once it reaches its melting point.

7 0

2 years ago

At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o

Damm [24]

Answer : The partial pressure of $H_2$ at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of $HBr$ = $1.0\times 10^{-2}atm$

The partial pressure of $H_2$ = $2.0\times 10^{-4}atm$

The partial pressure of $Br_2$ = $2.0\times 10^{-4}atm$

$K_p=4.2\times 10^{-9}$

The balanced equilibrium reaction is,

$2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)$

Initial pressure 1.0×10⁻² 2.0×10⁻⁴ 2.0×10⁻⁴

At eqm. (1.0×10⁻²-2p) (2.0×10⁻⁴+p) (2.0×10⁻⁴+p)

The expression of equilibrium constant $K_p$ for the reaction will be:

$K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}$

Now put all the values in this expression, we get :

$4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}$

$p=-1.99\times 10^{-4}$

The partial pressure of $H_2$ at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of $H_2$ at equilibrium is, 1.0 × 10⁻⁶

4 0

3 years ago

What information would you find on a WHIMIS label?

Nady [450]

Answer/explanation:

Product identifier – the brand name, chemical name, common name, generic name or trade name of the hazardous product.
Initial supplier identifier – the name, address and telephone number of either the Canadian manufacturer or the Canadian importer.

8 0

2 years ago

What's the formula of mercury (I) dioxonitrate III

steposvetlana [31]

<h2>Answer:</h2>

<h2>Explanation:</h2>

<h3>Formula of mercury ( 1 ) dioxonitrate 111 is Hg2 ( NO2 )2//</h3>

7 0

2 years ago