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tester [92]
3 years ago
10

Uestion 1

Chemistry
1 answer:
Bingel [31]3 years ago
5 0

Answer:

3.14 \times 10^{23} \text{ S atoms}

Explanation:

1. Calculate the moles of Al₂S₃

\text{Moles of Al$_{2}$S}_{3} = \text{26.1 g} \times \dfrac{\text{1 mol}}{\text{150.16 g}} = \text{0.1738 mol Al$_{2}$S}_{3}

2. Calculate the moles of S atoms

\text{Moles of S atoms} = \text{0.1738 mol Al$_{2}$S}_{3} \times \dfrac{\text{3 mol S atoms }}{\text{1 mol Al$_{2}$S}_{3}} = \text{0.5214 mol S atoms}

3. Calculate the number of S atoms

\text{No. of S atoms} = \text{0.5214 mol S atoms} \times \dfrac{6.022 \times 10^{23} \text{ S atoms}}{\text{1 mol S atoms}} = \mathbf{3.14 \times 10^{23}} \textbf{ S atoms}

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2. In Experiment SOL, you investigated the solubility of oxalic acid. Sodium oxalate, Na2C2O¬4, is the sodium salt of this acid.
AnnZ [28]

Answer:

Sodium oxalate is a basic salt. In water it can be dissolved and dissociated.

The oxalic acid in water has two dissociations.

Explanation:

Na2C2O4 ---> 2Na+   +  C2O4-2

Sodium oxalate is the conjugate base of a weak acid. In water this salt, dissociates completely giving rise to the sodium and oxalate ions. As Na+ comes from a strong base, in water it does not produce hydrolysis while oxalate does react in water, because it takes a proton from it and it generates a basic hydrolysis releasing OH-.

C2O4-2  + H2O ⇄  HC2O4-  +  OH-

In water the salt is basic.  The pH of an aqueous solution of this salt is basic, since OH- is generated.

The HC2O4- has a second hydrolisis, it takes another proton from water to form oxalic acid.

HC2O4-  +  H2O ⇄  H2C2O4  +  OH-

The oxalic acid acts as a weak acid, it can release 2 protons to water, to make oxalate (its conjugate base).

H2C2O4  + H2O ⇄ H3O+  + HC2O4-

HC2O4-  +  H2O ⇄  H3O+  C2O4-2

The  HC2O4-  acts as an ampholyte since it accepts and delivers protons simultaneously.

6 0
3 years ago
How do the test variables (independent variables) and outcome variables (dependent variables) in an experiment compare? A. The t
Ghella [55]

Answer:

I'm on the exact same queston

6 0
3 years ago
Read 2 more answers
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
What makes a compound a strong acid or base
SOVA2 [1]

If the acid is 100 percent dissociated in solutions of 1.0 M or less, it is called strong. Sulfuric acid is considered strong only in its first dissociation step; 100 percent dissociation isn't true as solutions become more concentrated.

3 0
3 years ago
Draw the condensed structural formula for the ester formed when each of the following reacts with methyl alcohol:
Vikki [24]

Answer:

1) HCOOCH3

2) CH3CH2COOCH3

3) CH3CH2CH2CH(CH3)COOCH3

Explanation:

In the reaction between an alcohol and a carboxylic acid, an ester and water are formed. It is analogous to the inorganic neutralization reaction but this reaction is called esterification in organic chemistry. Esters contain the general formula RCOOR where the RCOO moiety was obtained from the acid and the other R moiety was obtained from the alcohol. The -COOR shows the ester linkage in the molecule. The condensed structural formulas shown in the answer reflects these facts.

6 0
3 years ago
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