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2 years ago

W(OH)2 + 2 HCl → WCl2 + 2 H2O

Chemistry

1 answer:

Afina-wow [57]2 years ago

7 0

The amount of W(OH)2 needed would be 448.126 g

<h3>Stoichiometric calculation</h3>

From the equation of the reaction:

W(OH)2 + 2 HCl → WCl2 + 2 H2O

The mole ratio of W(OH)2 to HCl is 1:2

Mole of 150g HCl = 150/36.461

= 4.11 moles

Equivalent mole of W(OH)2 = 4.11/2

= 2.06 moles

Mass of 2.06 moles W(OH)2 = 2.06 x 217.855

= 448.188g

More on stoichiometric calculations can be found here: brainly.com/question/8062886

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One Ba2+ and two Br- ions will be there.

Answer is 3

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3 years ago

Please help i’ll mark brainliest

Lynna [10]

Answer:

Physical change

Explanation:

Chemical change is defined as the process in which new substance with different properties is formed from old substance and the process is irreversible while physical change is a reversible process and changes only physical properties of a substance.

The given experiment is an example of physical change because it is just the physical act of dissolving. The metal disk is dissolving in sodium acetate and the process is exothermic and the process is reversible as well.

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3 years ago

Hydrazine, , emits a large quantity of energy when it reacts with oxygen, which has led to hydrazine used as a fuel for rockets:

Tcecarenko [31]

Answer:

$1.25~mol~H_2O$ and $0.627~mol~N_2$

Explanation:

Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ( $N_2H_4$ ) and oxygen ( $O_2$ ). So, we can start with the reaction between these compounds:

$N_2H_4~+~O_2~->~N_2~+~H_2O$

Now we can balance the reaction:

$N_2H_4~+~O_2~->~N_2~+~2H_2O$

In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for $N_2H_4$ and 31.99 g/mol for $O_2$ ):

$20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4$

$20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2$

In the balanced reaction we have 1 mol for each reagent (the numbers in front of $O_2$ and $N_2H_4$ are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is $N_2H_4$ .

With this in mind, we can calculate the number of moles for each product. In the case of $N_2$ we have a 1:1 molar ratio (1 mol of $N_2$ is produced by 1 mol of $N_2H_4$ ), so:

$0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2$

We can follow the same logic for the other compound. In the case of $H_2O$ we have a 1:2 molar ratio (2 mol of $H_2O$ is produced by 1 mol of $N_2H_4$ ), so:

$0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O$

I hope it helps!

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4 years ago

1 mole solution of NaCl contains

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Answer:

If you dissolve 58.44g of NaCl in a final volume of 1 liter, you have made a 1M NaCl solution, a 1 molar solution.

Explanation:

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4 years ago

3) a) (2 points) Label the 1H NMR spectrum of adipic acid by identifying key resonances. Use letters (a, b, c, etc.) to match Hs

Bess [88]

Answer:

The label of spectrum is attached.

Explanation:

adipic acid is equal to:

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(1) (2) (3) (3) (2) (1)

The structure of adipic acid indicates that it has 3 equivalent carbons, therefore it should have 3 peaks in the spectrum:

The first peak at approximately 172 ppm is from carbon (1) COOH. The second peak is at approximately 23 ppm and is by carbon number (3) CH2. and the third peak is at approximately 37 ppm and is due to group number (2) CH2.

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3 years ago