Please help soon as possible!

B. The heat of reaction for the process described in (a) can be determined by

julsineya [31]

Answer: I believe the 1st and 3rd reactions are better obtained through reference sources and the 2nd and 4th are easiest and safest to measure in the laboratory.

Explanation:

I am also working on this Pre-lab right now, and I looked back at the first question to help get my answer. In the first question (a), it is noted that ammonia gas and gaseous hydrochloric acid are both potentially dangerous in gaseous form. I saw that both the 1st and 3rd reactions contained noxious gases (I knew this because there was a (g) in both of these reactions). Using the knowledge from the first question that the noxious gases were potentially dangerous, I assumed that those reactions were the ones that are better obtained through the reference sources. The 2nd and 4th reactions did not contain any noxious gases, so I assumed those ones were easiest and safest to measure in the laboratory. Hope this helps!

7 0

4 years ago

A pair of students found the temperature of 100. g of water to be 26.5°C. They then dissolved 8.89 g of AgNO3 in the water. When

bekas [8.4K]

Explanation:

Initial temperature of the water = $T_i=26.5^oC$

Final temperature of the water = $T_f=23.7^oC$

a) Change in temperature of the water,ΔT = $T_f-T_i$

$\Delta T=23.7^oC-26.5^oC=-2.8^oC$

Change in temperature of the water,ΔT is -2.8°°C.

b) Endothermic reaction : Reaction in which heat absorbed and the temperature if the surrounding is decreased.

Exothermic reaction : Reaction in which heat released and the temperature if the surrounding is increased.

On dissolving silver nitrate in water the temperature of the water decreased 2.8 degree Celsius which means that energy water was absorbed by solid silver nitrate to get dissolve in water. Hence, endothermic reaction.

(f) Mass of silver nitrate = 8.89 g

Moles of silver nitrate = $\frac{8.89 g}{170 g/mol}=0.05229 mol$

$AgNO_3(aq)\rightarrow Ag^+(aq)+NO_3^-(aq)$

1 mole of silver nitrate gives 1 mole of silver ion i.e. cation.

Then 0.05229 moles of silver nitrate will give:

$\frac{1}{1}\times 0.05229 mol=0.05229 mol$ silver ions.

0.05229 moles of cations are produced.

1 mole of silver nitrate gives 1 mole of nitrate ion i.e. anion .

Then 0.05229 moles of silver nitrate will give:

$\frac{1}{1}\times 0.05229 mol=0.05229 mol$ nitrate ions.

0.05229 moles of anion are produced.

5 0

3 years ago

A compound is 80.0% carbon and 20.0% hydrogen by mass. assume you have a 100.-g sample of this compound. the molar mass of the c

ch4aika [34]

Basis of the calculation: 100g

For Carbon:
Mass of carbon = (100 g)(0.80) = 80 g
Number of moles of carbon = (80 g)(1 mole / 12g) = 20/3

For Hydrogen:
Mass of hydrogen = (100 g)(0.20) = 20 g
Number of moles of hydrogen = (20 g)(1 mole / 1 g) = 20

Translating the answer to the formula of the substance,
C20/3H20

Dividing the answer,
CH3

The molar mass of the empirical formula is:
12 + 3 = 15 g/mol

Since, the molar mass given for the molecular formula is 30.069 g/mol, the molecular equation is,
C2H6

ANSWER: C2H6

4 0

4 years ago

Finding the density of solids requires a method of measuring the volume of the solid. If the solid has a regular geometric shape

Eduardwww [97]

Answer:

11 g/cm³

sorry if I got wrong

6 0

3 years ago

What is the ph of a 2.00 m hno2 solution if ka=5.00 e-6?

zysi [14]

Answer: -

2.5

Explanation: -

Initial [HNO₂] = 2.00 M

ka = 5.00 x 10 ⁻⁶

ICE table for HNO₂ =

[ $\left[\begin{array}{cccc}concentrations&HNO2&H+&NO2\\Initial&2.00&0&0\\Change&-x&+x&+x\\Equilibrium&2.00-x&x&x\end{array}\right]$

Ka = $\frac{x2}{2.00-x}$

5.00 x 10-6 = $\frac{x2}{2.00-x}$

x = 0.0032 M

pH = - log [H⁺]

= - log 0.0032

= 2.5

6 0

3 years ago