All categories

2 years ago

Use the balanced equation to solve the problem.

Chemistry

1 answer:

harkovskaia [24]2 years ago

7 0

Answer:

4.20 moles NF₃

Explanation:

To convert between moles of N₂ and NF₃, you need to use the mole-to-mole ratio from the balanced equation. This ratio consists of the coefficients of both molecules from the balanced equation. The molecule you are converting from (N₂) should be in the denominator of the ratio because this allows for the cancellation of units. The final answer should have 3 sig figs because the given value (2.10 moles) has 3 sig figs.

1 N₂ + 3 F₂ ---> 2 NF₃

2.10 moles N₂ 2 moles NF₃
--------------------- x --------------------- = 4.20 moles NF₃
1 mole N₂

You might be interested in

Calculate the wavelength (in nm) of the blue light emitted by a mercury lamp with a frequency of 6.32 × 1014 Hz.

tester [92]

Answer:

474 nm or 4.74 x 10^2 nm

Explanation:

c = λv

c (speed of light) = 2.998 x 10^8 m/s

λ = ?

v = 6.32 × 1014 Hz = 6.32 × 1014 1/s

2.998 x 10^8 m/s = (λ)(6.32 × 10^14 1/s)

λ = (2.998 x 10^8 m/s) / (6.32 × 10^14 1/s)

λ = 4.74 x 10^-7 m

λ = 4.74 x 10^-7 m x (1 x 10^9 nm/1 m) = 474 nm

7 0

2 years ago

The d-metals iron, copper, and manganese form cations with different oxidation states. For this reason, they are found in many o

choli [55]

Answer:

They have electrons in their 3d- and 4s-orbital for bond formation.

Explanation:

d- metals or transition metal are metal which form ion with partially filled d-orbital. Examples are iron and manganese.

The metals have 2 electrons in their 4s orbital. If only this is used for bonding, they will form compounds where they have oxidation State of +2 as seen in MnO.

If two 4s and one of 3d electrons are used, oxidation state of +3 is formed as seen in FeCl3.

If two 2s electron I used with two 3d electrons, compound with oxidation state of +4 is formed as seen in MnO2

4 0

3 years ago

Read 2 more answers

What is the pH of a solution with [H+]=6.2x10-9 M?

Serga [27]

Answer: 8.2

Explanation:

pH of a solution is - Log [ H+].

pH = - Log [ 6.2 x 10-9 M]

= 9 - 0.7924

pH = 8.24 approx 8.2

3 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

__Zn(s)+__CuSO4(aq)--->_______+________

stiks02 [169]

Answer:

Zn + CuSO4 —> ZnSO4 + Cu

Explanation:

Zn is higher than Cu in electrochemical series and so will displaces Cu in solution according to the equation:

Zn + CuSO4 —> ZnSO4 + Cu

5 0

2 years ago