Which elements are solid at room temperature

Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half r

Ugo [173]

Answer:

The ballance half reactions are:

Mg²⁺ + 2e⁻ → Mg

6OH⁻ + Si → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺ + 2e⁻ → Mg

Si → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺ + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺ + 4e⁻ → 2Mg

6OH⁻ + Si → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺ + 4e⁻ + 6OH⁻ + Si → 2Mg + SiO₃²⁻ + 4e⁻ + 3 H₂O

7 0

3 years ago

To answer this question, you will need to write the balanced equation and set up a BCA table. Using appropriate rounding rules,

Elenna [48]

Water decomposes when electrolyzed to produce hydrogen and oxygen gas. If 2.5 grams of water were decomposed 1.04 grams of oxygen will be formed.

BCA table:

2 $H_{2}$ O ⇒ $H_{2}$ + $O_{2}$

B 0.13 0 + 0

C -0.13 0.065 + 0.065

A 0 0.065

Explanation:

Balanced equation for water decomposition into hydrogen and oxygen gases

2 $H_{2}$ O ⇒ $H_{2}$ + $O_{2}$

B 0.13 0 + 0

C -0.13 0.065 + 0.065

A 0 0.065

Number of moles of water = $\frac{mass}{atomic mass of 1 mole}$

mass = 2.5 grams

atomic mass= 18 grams

number of moles can be known by putting the values in the formula,

n = $\frac{2.5}{18}$

= 0.13 moles

2 moles of water gives one mole of oxygen on decomposition

so, 0.13 moles of water will give x moles of oxygen on decompsition

$\frac{1}{2}$ = $\frac{x}{0.13}$

x = 0.065 moles of oxygen will be formed.

moles to gram will be calculated as

mass =number of moles x atomic mass

= 0.065 x 16

= 1.04 grams of oxygen.

7 0

3 years ago

_Fe2O3 + 2CO —> _Fe + _CO2

soldi70 [24.7K]

<h3>Answer:</h3>

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

<h3>Explanation:</h3>

Concept tested: Balancing of chemical equations

A chemical equation is balanced by putting appropriate coefficients on the products and reactants of the equation.
Balancing chemical equations ensures that chemical equations obey law of conservation of mass.
In this case; to balance the above equation we put the coefficients, 1, 3, 2, and 3 on the reactants and products.
Therefore; the balanced chemical equation for the reaction is;

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

8 0

3 years ago

Please someone help mee

Nikitich [7]

Answer:

13.9kj/mol

Explanation:

$h = 891 -(393.5 +2*241.8)=1768.1=13.9kj/mol$

8 0

3 years ago

Calculate the number of moles CuCl2 necessary to make 50 mL of a 0.15M solution

tekilochka [14]

Volume ⇒ 50 mL in liters : 50 / 1000 = 0.05 L

Molarity of solution ⇒ 0.15 M

Number of moles:

n = M * V

n = 0.15 * 0.05

n = 0.0075 moles of CuCl2

hope this helps!.

3 0

3 years ago