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3 years ago

Which determines the reactivity of an alkali metal?

Chemistry

2 answers:

astra-53 [7]3 years ago

3 0

Answer:

Its ability to lose electrons

Explanation:

The attraction from the positive nucleus to the negative electron is less. This makes it easier to remove the electron and makes the atom more reactive.

KiRa [710]3 years ago

3 0

Answer:

When it's able to lose electrons when it bonds with another substance.

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Which of the following does not form a diatomic molecule?

Murrr4er [49]

The correct answer among the choices given is option D. Copper does not form a diatomic molecule. A diatomic molecule has only two atoms. Diatomic molecules are molecule that are composed of two same atoms which are bonded covalently. An example are oxygen, it exist as O2 and chlorine, it exist as Cl2.

7 0

4 years ago

Read 2 more answers

How many atoms do sodium have?

nata0808 [166]

11, it has 11 atoms

7 0

3 years ago

A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

For NaI:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

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

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

4 0

3 years ago

Consider the chemical equation below.

Aleks [24]

I would say A but then again im not too sure so hope that makes it easier to somehow

6 0

3 years ago

Read 2 more answers

How many electrons in an atom can share the quantum numbers n = 4 and l = 3?

o-na [289]

Answer:

$\boxed{\text{14}}$

Explanation:

If l = 3, the electrons are in an f subshell.

The number of orbitals with a quantum number l is 2l + 1, so there

are 2×3 + 1 = 7 f orbitals.

Each orbital can hold two electrons, so the f subshell can hold 14 electrons.

$\boxed{\textbf{14 electrons}} \text{ can share the quantum numbers n = 4 and l = 3.}$

3 0

4 years ago