How many structures are possible for a trigonal bipyramidal molecule with a formula of ax3y2?

Tin has ten stable isotopes. The heaviest, 124Sn, makes up 5.80% of naturally occuring tin atoms. How many atoms of 124Sn are pr

matrenka [14]

The average atomic mass of Sn is 118.71 g/mol

the percentage of heaviest Sn is 5.80%

the given mass of Sn is 82g

The total moles of Sn will be = mass / atomic mass = 82/118.71=0.691

Total atoms of Sn in 82g = $6.023X10^{23}X0.691=4.16X10^{23}$

the percentage of heaviest Sn is 5.80%

So the total atoms of $Sn^{124}$ = 5.80% X $4.16X10^{23}$

Total atoms of $Sn^{124}$ = $2.41X10^{22}$ atoms

the mass of $Sn^{124}$ will be = $\frac{2.41X10^{22}X124}{6.023X10^{23}}=4.96g$

5 0

2 years ago

Consider the following hypothetical reaction: 2 P + Q â†’ 2 R + S The following mechanism is proposed for this reaction: P + P Q

aleksandr82 [10.1K]

Answer: the answer is option (D). k[P]²[Q]

Explanation:

first of all, let us consider the reaction from the question;

2P + Q → 2R + S

and the reaction mechanism for the above reaction given thus,

P + P ⇄ T (fast)

Q + T → R + U (slow)

U → R + S (fast)

we would be applying the Rate law to determine the mechanism.

The mechanism above is a three step process where the slowest step seen is the rate determining step. From this, we can see that this slow step involves an intermediate T as reactant and is expressed in terms of a starting substance P.

It is important to understand that laws based on experiment do not allow for intermediate concentration.

The mechanism steps for the reactions in the question are given below when we add them by cancelling the intermediates on the opposite side of the equations then we get the overall reaction equation.

adding this steps gives a final overall reaction reaction.

2P + Q ------------˃ 2R + S

Thus the rate equation is given as

Rate (R) = K[P]²[Q]

cheers, i hope this helps

3 0

2 years ago

1) How many aluminum atoms are there in 3.50 grams of Al2O3?

drek231 [11]

Answer:

Aluminium atoms = 4.13 *10^22 aluminium atoms

The correct answer is E

Explanation:

Step 1: Data given

Mass of Al2O3 = 3.50 grams

Molar mass of Al2O3 = 101.96 g/mol

Number of Avogadro = 6.022 * 10^23 /mol

Step 2: Calculate moles Al2O3

Moles Al2O3 = mass Al2O3 / molar mass Al2O3

Moles Al2O3 = 3.50 grams / 101.96 g/mol

Moles Al2O3 = 0.0343 moles

Step 3: Calculate moles Aluminium

In 1 mol Al2O3 we have 2 moles Al

in 0.0343 moles Al2O3 we have 2*0.0343 = 0.0686 moles Al

Step 4: Calculate aluminium atoms

Aluminium atoms = moles aluminium * Number of Avogadro

Aluminium atoms = 0.0686 * 6.022 * 10^23

Aluminium atoms = 4.13 *10^22 aluminium atoms

The correct answer is E

3 0

2 years ago

1. Suppose 0.7542 g of magnesium reacts with excess oxygen to form magnesium oxide as the only product, what would be the theore

Alina [70]

Answer :

(1) The theoretical yield of product, $MgO$ is, 1.257 grams.

(2) The percent yield of $MgO$ is, 64.13 %

(3) If the percent yield is calculated to be over 100% then there might be some impurity present in the desired product.

Solution : Given,

Mass of Mg = 0.7542 g

Molar mass of Mg = 24 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg.

$\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{0.7542g}{24g/mole}=0.03142moles$

Now we have to calculate the moles of $MgO$

The balanced chemical reaction is,

$2Mg(s)+O_2(g)\rightarrow 2MgO(s)$

From the reaction, we conclude that

As, 2 mole of $Mg$ react to give 2 mole of $MgO$

So, 0.03142 moles of $Mg$ react to give 0.03142 moles of $MgO$

Now we have to calculate the mass of $MgO$

$\text{ Mass of }MgO=\text{ Moles of }MgO\times \text{ Molar mass of }MgO$

$\text{ Mass of }MgO=(0.03142moles)\times (40g/mole)=1.257g$

Theoretical yield of $MgO$ = 1.257 g

Experimental yield of $MgO$ = 0.8922 g

Now we have to calculate the percent yield of $MgO$

$\% \text{ yield of }MgO=\frac{\text{ Experimental yield of }MgO}{\text{ Theretical yield of }MgO}\times 100$

$\% \text{ yield of }MgO=\frac{0.8922g}{1.257g}\times 100=70.97\%$

Therefore, the percent yield of $MgO$ is, 70.97 %

If the percent yield is calculated to be over 100% then the product would be greater than 1.257 g which indicates that there might be some impurity present in the desired product.

4 0

2 years ago

A pure gold bar is made of 19.55 mol of gold. What is the mass of the bar in grams

Law Incorporation [45]

<span>(19.55 mol Au) / ( 1 ) x (196.97 g Au) / ( 1 mol Au) =
19.55 x 196.97 =
3850.76 g Au

I hope this helps you and have a great day!! :)
</span>

4 0

1 year ago

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