Describing Nuclear Decay

Which greenhouse gas is emitted when people burn fossil fuels for energy?

Volgvan

It would be carbon dioxide because carbon dioxide makes up the vast majority of greenhouse gas .

5 0

3 years ago

The molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution is ________.

Nonamiya [84]

Answer:

1.86 M

Explanation:

From the question given above, the following data were obtained:

Mass of sucrose (C12H22O11) = 22.5 g

Volume of solution = 35.5 mL

Molarity of solution =?

Next, we shall determine the number of mole in 22.5 g of sucrose (C12H22O11). This can be obtained as follow:

Mass of sucrose (C12H22O11) = 22.5 g

Molar mass of C12H22O11 = (12×12) + (22×1) + (16×11)

= 144 + 22 + 176

= 342 g/mol

Mole of C12H22O11 =?

Mole = mass /Molar mass

Mole of C12H22O11 = 22.5 /342

Mole of sucrose (C12H22O11) = 0.066 mole

Next, we shall convert 35.5 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

35.5 mL = 35.5 mL × 1 L / 1000 mL

35.5 mL = 0.0355 L

Thus, 35.5 mL is equivalent to 0.0355 L.

Finally, we shall determine the molarity of the solution as follow:

Mole of sucrose (C12H22O11) = 0.066 mole

Volume of solution = 0.0355 L.

Molarity of solution =?

Molarity = mole /Volume

Molarity of solution = 0.066/0.0355

Molarity of solution = 1.86 M

Therefore, the molarity of the solution is 1.86 M.

8 0

3 years ago

Which of the following is true for an excess reactant?

kupik [55]

Answer:

C) It is the reactant that is left over after the reaction stops.

Explanation:

The excess reactant is the reactant that is left over after the reaction stops. The extent of the reaction is not determined by this reactant.

A limiting reactant is a reactant that is in short supply within a given reaction.

Such reactants determines the extent of chemical reaction.

Limiting reactants are used up in a chemical reaction.
The excess reactants remains unchanged after the reaction.

8 0

3 years ago

Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

Answer: The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

Explanation:

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

$HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)$

Initial: 0.025 0.025

At eqllm: 0.025-x 0.025+x x

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

We know that:

$Ka_2\text{ for }H_2SO_4=0.01$

Putting values in above equation, we get:

$0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

4 0

3 years ago

Brain cells, blood cells, muscle cells These cells are an example of:?

TEA [102]

Answer:

Somatic Cells

Explanation:

Somatic cells are basically all bodily cells

4 0

3 years ago

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