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3 years ago

If an element has 43 electrons, 56 neutrons, and 43 protons, what is its approximate atomic mass?

1 answer:

4 0

Approximately 99 because there is the same amount of electrons and the only things that go into the nucleus is protons and neutrons.

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If the solvent is a liquid, then the solute:

Lelechka [254]

Answer:

must be either a gas or liquid

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3 years ago

What effect does high temperature have on radish growth size?

Juliette [100K]

Radishes are a popular root vegetable known scientifically as Raphanus raphanistrum. Radishes come in a variety of colors, including black, red, purple, and white. They have been harvested for thousands of years.

3 0

3 years ago

Read 2 more answers

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

What is the coefficient for water after the equation is balanced?

Pie

Answer:

Explanation:

6 0

2 years ago

Isotopes that decay with a nuclear break-up and emit a significant amount of energy are said to be ________.

podryga [215]

Answer to this is Radioactive isotopes.

Isotopes are the species of the same element having different atomic masses that means the number of protons remains the same but number of neutrons do differ. For example $_{1}^{2}\textrm{H}$ and $_{1}^{3}\textrm{H}$ are the two isotopes of Hydrogen ( $_{1}^{1}\textrm{H}$ ).

Radioactive isotopes are the isotopes which release some kind of energy in the form of alpha particles, beta particles or gamma radiation. Examples of each of the decay processes are :

Alpha Decay: In this decay one alpha particle having atomic mass 4 and atomic number 2 or we can say a He molecule will come out. $_{Z}^{A}\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_{2}^{4}\alpha$

Beta Decay: In this decay a $\beta$ particle is emitted increasing the atomic number of the reactant by 1 unit.

$_{Z}^{A}\textrm{X}\rightarrow _{Z+1}^{A}\textrm{Y}+_{-1}^{0}\beta$

Gamma Radiation: In this type of reaction only radiation is emitted out which does not change the original molecule.

$_{Z}^{A}\textrm{X}\rightarrow _{Z}^{A}\textrm{X}+\gamma\text{ radiation}$

3 0

3 years ago