Which of the following contains the GREATEST number of atoms of argon gas?

Which types of changes must follow the law of conservation of mass? (2 points)

Oksana_A [137]

Both Physical and chemical changes.

4 0

3 years ago

An atom of which element has the greatest attraction for the electrons in a bond with a hydrogen atom?

Nadusha1986 [10]

The question can be changed into a new form:
Which element has the most negative electron affinity, or attraction for electrons? halogens have the highest electron affinities, and thus are more attracted to the electrons in the Hydrogen atom than any element in their respective periods.
In this case all the following choices are in the same period, thus Cl or Chlorine is the answer as it is a halogen.

7 0

3 years ago

a sample of solid is decomposed and found to contain 6.52g of potassium, 4.34 g of chromium and 5.34 of oxygen, what is the empi

Fynjy0 [20]

Answer:

K₂CrO₅

Explanation:

The empirical formula is the simplest formula of a compound. To find the empirical formula, we follow the procedure below:

Elements Potassium Chromium Oxygen

Mass 6.52 4.34 5.34

Molar mass 39 60 16

Number of moles 6.52/39 4.34/60 5.34/16

0.167 0.072 0.333

Divide through by

the smallest 0.167/0.072 0.072/0.072 0.333/0.072

2.3 1 4.6

2 1 5

Empirical formula K₂CrO₅

5 0

3 years ago

Please help me. How do I do ideal gas law?

Tcecarenko [31]

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

4 0

3 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

3 years ago