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Y_Kistochka [10]

3 years ago

10

the general name for a substance added to a reaction that affects the rate but is not consumed in the reaction is called a_____.

complex constituent reactant catalyst

2 answers:

stiks02 [169]3 years ago

8 0

It's complex, google :)

cluponka [151]3 years ago

4 0

Answer:

complex

Explanation:

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3 years ago

Trace amounts of sulfur (S) in coal are burned in the presence of diatomic oxygen (O2) to form sulfur dioxide (SO2). Determine t

Mashcka [7]

Answer:

0.99 kg O₂

1.9 kg SO₂

Explanation:

Let's consider the reaction between sulfur and oxygen to form sulfur dioxide.

S + O₂ → SO₂

The mass ratio of S to O₂ is 32.07:32.00. The mass of oxygen required to react with 1 kg of sulfur is:

1 kg S × (32.00 kg O₂/32.07 kg S) = 0.998 kg O₂

The mass ratio of S to SO₂ is 32.07:64.07. The mass of sulfur dioxide formed when 1 kg of sulfur is burned is:

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3 years ago

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UNO [17]

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4 0

2 years ago

The specific gravity of water is 1.0. True False

Zielflug [23.3K]

The specific gravity of water is 1.0 True

5 0

2 years ago

Be sure to answer all parts.

MrRissso [65]

Answer: The molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

Explanation:

Molarity is the number of moles of a substance present in liter of a solution.

And, moles is the mass of a substance divided by its molar mass.

(a) Moles of ethanol (molar mass = 46 g/mol) is as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{28.5 g}{46 g/mol}\\= 0.619 mol$

Now, molarity of ethanol solution is as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.619 mol}{4.50 \times 10^{2} \times 10^{-3}L}\\= 1.38 M$

(b) Moles of sucrose (molar mass = 342.3 g/mol) is as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{21.6 g}{342.3 g/mol}\\= 0.063 mol$

Now, molarity of sucrose solution is as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.063 mol}{0.067 L} (1 mL = 0.001 L)\\= 0.94 M$

(c) Moles of sodium chloride (molar mass = 58.44 g/mol) are as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{6.65 g}{58.44 g/mol}\\= 0.114 mol$

Now, molarity of sodium chloride solution is as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.114 mol}{0.0962 L}\\= 1.182 M$

Thus, we can conclude that the molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

4 0

2 years ago