NA + 1.82 x nA = 2.0 x 10^-2. How do you solve for nA?

Margaret [11]

Answer: The structure of an atom, theoretically consisting of a positively charged nucleus surrounded and neutralized by negatively charged electrons revolving in orbits at varying distances from the nucleus, the constitution of the nucleus and the arrangement of the electrons differing with various chemical elements.

:) I hope this helped! :)

4 0

4 years ago

An example of a double replacement chemical reaction takes place in batteries. In an alkaline battery, Manganese oxide reacts wi

ozzi

Answer:

2MnO2 + H2O => Mn2O3 + 2OH is the correct balanced equation.

6 0

3 years ago

I need help ASAP!

MaRussiya [10]

Calcium reacts gently with water to give hydrogen and calcium hydroxide, which is only slightly soluble, thus slows down the reaction.

It will be assumed that hydrochloric acid used is a dilute aqueous solution.

However, calcium reacts with hydrochloric acid to give calcium chloride which is readily soluble in water, and hydrogen, being a typical reaction of relatively active metals with acids.
Ca(s) + 2HCl(aq) -> CaCl2(aq) +H2(g) ↑ + heat

The clues that it is a chemical reaction could be:
- formation of a new substance, gaseous hydrogen
- disappearance of a metallic solid in the solution
- heat formed during the vigorous reaction.

As silver is below hydrogen in the electrochemical series, it will not be expected to react with dilute hydrocloric acid. (however, it dissolves in oxidizing acid such as nitric acid, but not displacing hydrogen as a product).

8 0

3 years ago

A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g of hydrogen. Find the formula of the compound

Karolina [17]

Answer:

The empirical formula of the compound is $C_{0.504}HO_{1.008}$ .

Explanation:

We need to determine the empirical formula in its simplest form, where hydrogen ( $H$ ) is scaled up to a mole, since it has the molar mass, and both carbon ( $C$ ) and oxygen ( $O$ ) are also scaled up in the same magnitude. The empirical formula is of the form:

$C_{x}HO_{y}$

Where $x$ , $y$ are the number of moles of the carbon and oxygen, respectively.

The scale factor ( $r$ ), no unit, is calculated by the following formula:

$r = \frac{M_{H}}{m_{H}}$ (1)

Where:

$m_{H}$ - Mass of hydrogen, in grams.

$M_{H}$ - Molar mass of hydrogen, in grams per mole.

If we know that $M_{H} = 1.008\,\frac{g}{mol}$ and $m_{H} = 0.2\,g$ , then the scale factor is:

$r = \frac{1.008}{0.2}$

$r = 5.04$

The molar masses of carbon ( $M_{C}$ ) and oxygen ( $M_{O}$ ) are $12.011\,\frac{g}{mol}$ and $15.999\,\frac{g}{mol}$ , then, the respective numbers of moles are: ( $r = 5.04$ , $m_{C} = 1.2\,g$ , $m_{O} = 3.2\,g$ )