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andrew11 [14]
2 years ago
7

A student wants to measure the heat of neutralization reaction. To do this they mixed in a coffer cup calorimeter 58.0 mL of 0.5

0M HCl solution and 42.0 mL of 0.60M NaOH solution both at 22.0oC initially. They then continued to measure the temperature of the mixture for 10 minutes. What is the SURROUNDING of this reaction?
Chemistry
1 answer:
adoni [48]2 years ago
6 0

Answer:

Explanation:

Determine the molar masses to the nearest whole number.

NaoH

First, use the molar masses of N and O to determine the molar mass of N2O5.

2 mol N 14.0 g N

1 mol N

× + 5 mol O 16.0 g O

1 mol O

× -1 = 108.0 g mol ⋅

Next, use the molar mass to convert the given moles into mass.

3.24 mol N O2 5 × 2 5

2 5

108.0 g N O

1 mol N O 2 5 = 350. g N O

b) How many moles of N2O5 are present in a 12.7-g sample of N2O5?

Use the molar mass determined in Part A to convert from mass to moles.

12.7 g N O2 5 × 2 5

2 5

1 mol N O

108.0 g N O 2 5

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Answer:

Hg^0+2OH^-\rightarrow Hg^{2+}O+H_2O+2e^-

Explanation:

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In this case, mercury (II) oxide (HgO) is obtained via the reaction:

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Hg^0+2OH^-\rightarrow Hg^{2+}O+H_2O+2e^-

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2 years ago
What is the overall charge of the compound frbr
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Answer:

Zero  

Explanation:

FrBr is an ionic compound .

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Calculate the molarity of sodium chloride in a half-normal saline solution (0.45% NaCl). The molar mass of NaCl is
Rus_ich [418]

Answer:

0.077 M

Explanation:

Data Given :

The concentration of half normal (NaCl) saline = 0.45g / 100 g

So,

Volume of Solution = 100 g = 100 mL

Volume of Solution in Liter = 100 mL / 1000

Volume of Solution = 0.1 L

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Molarity:

Molarity is the representation of the solution. It is amount of solute in moles per liter of solution and represented by M

Formula used for Molarity

                M = moles of solute / Liter of solution . . . . . . . . . . (1)

Now to find number of moles of Nacl

                no. of moles of NaCl = mass of NaCl / molar mass

                no. of moles of NaCl = 0.45g / 58.44 g/mol

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Put values in the eq (1)

                  M = moles of solute / Liter of solution . . . . . . . . . . (1)

                  M = 0.0077 g / 0.1 L

                  M = 0.077 M

So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M

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