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4 years ago

Can someone help me with this please

Chemistry

1 answer:

never [62]4 years ago

3 0

Answer: B. 1:2

Explanation: Beryllium and chlorine forms a binary ionic compound. Ionic compound is formed when a metal loses its electrons to a receiving non metal. Beryllium (metal) has two valence electrons while chlorine (nonmetal) has seven valence electrons, and so a beryllium atom has to give out its two valence electrons to attain a duplet stable structure while a chlorine atom will gain one electron to attain its stable octet structure. In the reaction between beryllium and chlorine, two atoms of chlorine have to accept the two electrons from one beryllium atom to attain their stable octet structure.

The formula of the compound formed is BeCl2.

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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

A sealed helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 293 K is moved. What volume will the bal

bixtya [17]

Answer:

6.48 L

Explanation:

From the question,

Applying

PV/T = P'V'/T'......................... Equation 1

P = initial pressure of the helium balloon, V = Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.

make V' the subject of the equation

V' = PVT'/P'T......................... Equation 2

Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm

Substitute these values into equation 2

V' = (4.5×1×253)/(0.6×293)

V' = 1138.5/175.8

V' = 6.48 L

8 0

3 years ago

If the gas in the piston above has a volume of 20.0 L at a temperature of 25 C what is the volume of that gas when it is heated

dexar [7]

Answer:

15.98 L

Explanation:

First, you need to find T1, T2, V1 and V2.

T1 = 25 C = 298.15 K (25C + 273.15K)

T2 = 100 C = 373.15 K (100C + 273.15K)

V1 = 20. L

V2 = ? (we are trying to find)

Next, rearrange to fit the formula

V2 = V1 x T1 / T2

Next, fill in with our numbers

V2 = 20. L x 298.15 K / 373.15 K

Do the math and you should get...

15.98 L

- If you need more help or futher explanation please let me know. I would be glad to help!

8 0

3 years ago

Read 2 more answers

Silver is a white metal that is an excellent conductor. Silver tarnishes when exposed to air and light. The density of silver is

Stells [14]

Answer:

c tarnishes in air

Explanation:

After silver has been exposed to air that contains sulphur gases, discoloration would occur. there would be darkening that is caused by the reaction with gases.When any silver object tarnishes, it brings about a disfiguring of that object. Hydrogen sulphide would be needed for this to happen. silver sulphide is black and a if a thin layer should form on any surface, it ill darken it. This is what we refer to as tarnishing.

3 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST

OverLord2011 [107]

Answer:

butylphenyl ether.

Explanation:

The formula of the compound is:

CH₃ - CH₂ - CH₂ - CH₂ - O - C₆H₅

1. The functional group is of the kind R - O - R', i.e. two alkyl groups each attached to one end of the oxygen atom. That means that the compound is an ether.

2. One group attached to the oxygen group is CH₃ - CH₂ - CH₂ - CH₂ - which has 4 carbons and is named butyl group.

3. The other group attached to the oxygen atom is C₆H₅ - which is derived from ciclohexane as is known as phenyl group.

4. Using the rule of naming the subtituents in alphabetical order, you name butyl first and phenyl second, so it is butylphenyl ether.

4 0

3 years ago