All categories

2 years ago

Write correctly a balanced equation for the following "word" equation: Red lead → lead monoxide + oxygen

Chemistry

2 answers:

user100 [1]2 years ago

7 0

Answer:

Red lead_lead monoxide +oxygen

Nat2105 [25]2 years ago

6 0

Answer:

nothing haha because I don't know what is the answer

You might be interested in

A- is a weak base. which equilibrium corresponds to the equilibrium constant ka for ha?

vekshin1

Answer:

$K_a=\frac{[H^+][A^-]}{[HA]}$

Explanation:

ka is defined as the dissociation constant of an acid. It is defined as the ratio of concentration of products to the concentration of reactants.

For the dissociation of weak acid, the chemical equation follows:

$HA\rightleftharpoons H^++A^-$

The equilibrium constant is defined by the equilibrium concentration of products over reactants:

$K_a=\frac{[H^+][A^-]}{[HA]}$

5 0

3 years ago

When butane burns completely, only water and carbon dioxide gas are produced. If 11.6 g of butane and 40.0 L of oxygen at 22.0o

Angelina_Jolie [31]

19.7 litre volume of carbon dioxide gas at 22.0o C and 102 kPa can be collected over water.

<h3>What is vapour pressure?</h3>

Vapour pressure is a measure of the tendency of a material to change into the gaseous or vapour state, and it increases with temperature.

Moles of Butane = mass in grams / molar mass = 11.6 / 58.12 = 0.2

Volume of $O_2$ (V) = 40 liter

Temperature (T) = 22°C = 22 + 273 = 295 K

Pressure (P) = 102 kPa = 102 / 101.325 = 1.007 atm

Moles of $O_2$ (n) can be calculated by ideal gas equation.

PV = nRT

n = 1.007 40 ÷ 0.0821 295 = 1.663

Balanced chemical reaction;

2 $C_4H_10$ + 13 $O_2$ ---> 8 $CO_2$ + 10 $H_2O$

From reaction;

13 moles $O_2$ require 2 moles $C_4H_10$

So, 1.663 moles $O_2$ will require = 2 x 1.663 ÷13 = 0.256 moles of $C_4H_10$

Thus $C_4H_10$ is a limiting reagent. So it will drive the yield of $CO_2$ .

Moles of $CO_2$ produced = (8/2) 0.2 = 0.8 moles

Pressure of $CO_2$ (P) = 102 - 2.24 = 99.76 kPa = 99.76 ÷ 101.325 = 0.985 atm

Applying the ideal gas equation for $CO_2$ ,

PV = nRT

0.985 V = 0.8 0.0821 x 295

V = 19.7 liter

The volume of $CO_2$ produced = 19.7 liter.

Learn more about the vapour pressure here:

brainly.com/question/25699778

#SPJ1

5 0

1 year ago

Write the formula for potassium oxide. why do you not need prefixes in the name

Molodets [167]

Potassium oxide: K₂O.

There's no need for prefixes since K₂O is an ionic compound.

<h3>Explanation</h3>

Find the two elements on a periodic table:

Potassium- K- on the left end of period four.
Oxygen- O- near the right end of periodic two.

Elements on the bottom-left corner of the periodic table are metals. Those on the top-right corner are nonmetals.

Potassium is a metal,
Oxygen is a nonmetal.

A metal and a nonmetal combine to form an ionic compound. Potassium oxide is likely to be an ionic compound. It contains two types of ions:

Potassium ions: Potassium is group 1 of the periodic table. It is an alkaline metal. Like other alkaline metals such as sodium Na, potassium K tends to lose one electron and form ions of charge +1 in compounds. The ion would be K⁺.

Oxide ions from oxygen: Oxygen is the second most electronegative element on the periodic table. It tends to gain two electrons and form the oxide ion $\text{O}^{2-}$ when it combines with metals.

The two types of ions carry opposite charges. They shall pair up at a certain ratio such that they balance the charge on each other. The charge on each $\text{O}^{2-}$ ion is twice that on a $\text{K}^{+}$ ion. Each $\text{K}^{+}$ would pair up with two $\text{O}^{2-}$ . Hence the subscript in the formula: $\text{K}_{\bf 2}\text{O}$ .

There are two classes of compounds:

Covalent compounds, which need prefixes, and
Ionic compounds, which need no prefix.

Prefixes are needed only in covalent compounds. For instance in the covalent compound carbon dioxide $\text{CO}_2$ , the prefix di- indicates that there are two oxygen atoms in the formula $\text{CO}_2$ . However, there's no need for prefix in ionic compounds such as $\text{K}_2\text{O}$ .