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jolli1 [7]
3 years ago
13

PLEASE HELPPP!!!!!!!!!!!!!!!!!!!!

Chemistry
2 answers:
riadik2000 [5.3K]3 years ago
6 0
Similarities. very useful
differences. coal is a pullutant. while womd is not. i guess
arlik [135]3 years ago
4 0
Similarities: both natural differences: the work differently
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How many molecules are found in 90 grams of Al(OH)3?
jarptica [38.1K]

Answer:

6.95 x 10²³ molecules/particles

Explanation:

First we need to find the total Empirical Mass. We can find this by adding each element's mass together.

Al = 26.982,

O = 15.999

H = 1.008

26.982 + 3(15.999) + 3(1.008) = 78.003.

Now we divide by the mass given (90 grams).

90/78.003 = 1.153801777.

We then take that number and multiply it by avogadro's number (6.022 x 10²³)

1.153801777 x avogadro's number = 6.95 x 10²³

6 0
3 years ago
When you combine 50.0 mL of 0.100 M AgNO3 with 50.0 mL of 0.100 M HCl in a coffee-cup calorimeter, the temperature changes from
RideAnS [48]

Answer : The enthalpy of reaction (\Delta H_{rxn}) is, 67.716 KJ/mole

Explanation :

First we have to calculate the moles of AgNO_3 and HCl.

\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole

\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole

Now we have to calculate the moles of AgCl formed.

The balanced chemical reaction will be,

AgNO_3(aq)+HCl(aq)\rightarrow AgCl(s)+HNO_3(aq)

As, 1 mole of AgNO_3 react with 1 mole of HCl to give 1 mole of AgCl

So, 0.005 mole of AgNO_3 react with 0.005 mole of HCl to give 1 mole of AgCl

The moles of AgCl formed  = 0.005 mole

Total volume of the solution = 50.0 ml + 50.0 ml = 100.0 ml

Now we have to calculate the mass of solution.

Mass of the solution = Density of the solution × Volume of the solution

Mass of the solution = 1.00 g/ml × 100.0 ml = 100 g

Now we have to calculate the heat.

q=m\times C\Delta T=m\times C \times (T_2-T_1)

where,

q = heat

C = specific heat capacity = 4.18J/g^oC

m = mass = 100 g

T_2 = final temperature = 24.21^oC

T_1 = initial temperature = 23.40^oC

Now put all the given values in the above expression, we get:

q=100g\times (4.18J/g^oC)\times (24.21-23.40)^oC

q=338.58J

Now  we have to calculate the enthalpy of the reaction.

\Delta H_{rxn}=\frac{q}{n}

where,

\Delta H_{rxn} = enthalpy of reaction = ?

q = heat of reaction = 338.58 J

n = moles of reaction = 0.005 mole

Now put all the given values in above expression, we get:

\Delta H_{rxn}=\frac{338.58J}{0.005mole}=6771.6J/mole=67.716KJ/mole

Conversion used : (1 KJ = 1000 J)

Therefore, the enthalpy of reaction (\Delta H_{rxn}) is, 67.716 KJ/mole

4 0
3 years ago
ACTUAL YIELD VS THEORETICAL YIELD?
lawyer [7]

Actual yield over theoretical yield, then multiply by 100

6 0
3 years ago
Although 12 is the stable form of iodine at low
kompoz [17]

Answer:

The first option----A

6 0
3 years ago
What is true of a scientific law?
Minchanka [31]

Answer: d

Explanation:

5 0
3 years ago
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