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artcher [175]
3 years ago
12

Determine the percent water in MgSO4*7H20?

Chemistry
1 answer:
Semmy [17]3 years ago
5 0

Answer:

\% H_2O=51.2\%

Explanation:

Hello!

In this case, since the percent water is computed by dividing the amount of water by the total mass of the hydrate; we infer we first need the molar mass of water and that of the hydrate as shown below:

MM_{MgSO_4* 7H_2O}=120.36 g/mol+7*18.02g/mol\\\\MM_{MgSO_4* 7H_2O}=246.5g/mol

Thus, the percent water is:

\% H_2O=\frac{7*MM_{H_2O}}{MM_{MgSO_4* 7H_2O}} *100\%\\\\

So we plug in to obtain:

\% H_2O=\frac{7*18.02}{246.5} *100\%\\\\\% H_2O=51.2\%

Best regards!

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Suppose the price of gasoline increases 10% and quantity of gasoline demanded in Orlando drops 5% per day. Demand for gasoline i
grin007 [14]

Answer:

The correct answer is;

Demand for gasoline in Orlando is price inelastic.

Explanation:

The elasticity is the degree of response to a change in price or quantity supplied to the the quantity demanded. An elastic demand responds positively to change in price, while an inelastic demand means that when there is a price increase, the quantity demanded remains the same and where there is a drop in price the quantity demanded remains constant.

If a small change in price results in a large change in demand then the good is said to be price elastic

In the question the price increases by 10% while the quantity demanded drops 5 % daily. Therefore it is price inelastic

5 0
3 years ago
A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0
3 years ago
I need help 1-4 plz
LekaFEV [45]
The answer is -3, if you are asking for that
5 0
3 years ago
What two categorical ways of thinking about matter are discussed in the passage?
Mamont248 [21]

Answer:

There's no passage but I can say that matter are that ''thing'' which extend through all Universe, matter is everything. It can be seen or felt as liquid, gas or solid. This changes of its state depend on the amount of energy that is involved.

So, here is used categorical thinking because is a abstract concept that is hard to understand sometimes. It's abstract when we say that matter is all, not  just what we see, but even what we don't see like ''Dark Matter'', which is the majority in the Universe.

4 0
3 years ago
What is the percent of hydrogen peroxide in the solution when 5.02g of H2O2 solution reacts to produce 0.153g of H2O2? 2H2O2 (l)
Murljashka [212]

<u>Given:</u>

Mass of H2O2 solution = 5.02 g

Mass of H2O2 = 0.153 g

<u>To determine: </u>

The % H2O2 in solution

<u>Explanation:</u>

Chemical reaction-

2H2O2(l) → 2H2O(l) + O2(g)

Mass % of a substance in a solution = (Mass of the substance/Mass of solution) * 100

In this case

% H2O2 = (Mass H2O2/Mass of solution)* 100 = (0.153/5.02)*100 = 3.05%

Ans: % H2O2 in the solution = 3.05%

5 0
3 years ago
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