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kirza4 [7]
3 years ago
8

In an experiment Teresa's measures 15.5 mL of water she must have used a

Chemistry
1 answer:
kotykmax [81]3 years ago
3 0

Answer:

transfer pipet that had markings every 0.1 mL.

Explanation:

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Genes are located in which part of a cell?
ddd [48]

Answer:A

Explanation:

7 0
3 years ago
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch
mamaluj [8]

Answer:

Explanation:

From the given information:

The equation for the reaction can be represented as:

2SO_2 + O_2 \to 2SO_3

The I.C.E table can be represented as:

                     2SO₂              O₂                   2SO₃

Initial:             14                  2.6                     0

Change:        -2x                -x                      +2x

Equilibrium:   14 - 2x          2.6 - x                2x

However, Since the amount of sulfur trioxide gas to be 1.6 mol.

SO₃ = 2x,

then x = 1.6/2

x = 0.8 mol

For 2SO₂; we have 14 - 2x

= 14 - 2(0.8)

= 14 - 1.6

= 12.4 mol

For O₂; we have 2.6 - x

= 2.6 - 1.6

= 1.0 mol

Thus;

[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,

[O₂] = 1/50 = 0.02 M ,  

[SO₃] = 1.6/50 = 0.032 M

Kc = [SO₃]² / [SO₂]² [O₂]

= ( 0.032²) / ( 0.248² x 0.02)

= 0.8325

Recall that; the equilibrium constant for the reaction 2SO_2 + O_2 \to 2SO_3 = 0.8325;

If we want to find:

SO_2 + \dfrac{1}{2}O_2 \to SO_3

Then:

K_c = (0.8325)^{1/2}

\mathbf{K_c = 0.912}

Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.

7 0
3 years ago
Chemists investigated an unknown substance and found it to have the following characteristics:
juin [17]

D the substance is a homogeneous mixture

8 0
3 years ago
In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2
valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

\dfrac{80}{17}=4.706

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

\tt \dfrac{120}{32}=3.75

Mol ratio of reactants(to find limiting reatants) :

\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

\tt \dfrac{6}{5}\times 3.75=4.5

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%

6 0
3 years ago
3. Given the following equation:
miskamm [114]

Answer:

4.767 grams of KCl are produced from 2.50 g of K and excess Cl2

Explanation:

The balanced equation is

2 K+ Cl2 --->2 KCI

Here the limiting agent is K. Hence, the amount of KCl will be calculated as per the mass of 2.50 gram of K

Mass of one atom/mole of potassium is 39.098 grams

Number of moles is 2.5 grams = \frac{2.5}{39.098} = 0.064

So, 2 moles of K produces 2 moles of KCL

0.064 moles of K will produces 0.064 moles of KCl

Mass of one molecule of KCl is 74.5513 g/mol

Mass of 0.064 moles of KCl is 4.767 grams

4 0
2 years ago
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