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3 years ago

Given the following reaction: 2H2(g) + O2(g) à 2H2O(g) answer the following questions:

Chemistry

1 answer:

lora16 [44]3 years ago

6 0

Answer:

Explanation:

2 moles hydrogen reacts with one mole of oxygen to give 2 moles of water.

a ) rate of consumption of hydrogen ( moles per second) is twice the rate of consumption of oxygen .

b ) rate of formation of water ( moles per second ) is twice the rate of consumption of oxygen

c ) rate of formation of water ( moles per second ) is equal to the rate of consumption of hydrogen.

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Cassini was the fourth space probe to visit Saturn and the first to enter its orbit.

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3 years ago

How many atoms are in 19.6 g of Sodium?

Hitman42 [59]

Answer:

Number of atoms = 5.1345 x 10²³

Explanation:

Given:

Mass of sodium = 19.6

Find:

Number of atoms

Computation:

We know that atomic mass of sodium is 22.98

So,

Number of atoms = [Mass of sodium/atomic mass of sodium]Avogadro's number

Number of atoms = [19.6 / 22.98]6.02 x 10²³

Number of atoms = 5.1345 x 10²³

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3 years ago

Help solve this page please

AURORKA [14]

Answer:

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3 years ago

6)Explain why a 0.010 M solution of hydrochloric acid, HCl has a pH of 2.00 but a 0.010 M solution of

ExtremeBDS [4]

0.010 M solution of hydrochloric acid, HCl has a pH of 2.00 but a 0.010 M solution of acetic acid, HCH₃COO, is 3.37 is because HCl dissociates more.

<h3>Dissociation</h3>

This is the separation of ions of an ionic compound when it dissolves.

<h3>Hydrogen ion concentration of HCl</h3>

Since 0.010 M solution of hydrochloric acid, HCl has a pH of 2.00 its hydrogen ion concentration is gotten from

pH = -log[H⁺]

So,

$[H^{+} ] = 10^{-pH}$

So, [H⁺] = 10⁻²

<h3>Hydrogen ion concentration of HCH₃COO</h3>

Also, since 0.010 M solution of acetic acid, HCH₃COO, has a pH of 3.37, its hydrogen ion concentration is gotten from

pH = -log[H⁺]

So,

$[H^{+} ] = 10^{-pH}$

So, [H⁺] = 10⁻³°³⁷ = 0.000427 = 4.27 × 10⁻⁴

Since the hydrogen ion concentration of HCl is 10⁻² and the hydrogen ion concentration of HCH₃COO is 4.27 × 10⁻⁴, we see that HCl dissociates more than HCH₃COO.

So, 0.010 M solution of hydrochloric acid, HCl has a pH of 2.00 but a 0.010 M solution of acetic acid, HCH₃COO, is 3.37 is because HCl dissociates more.

Learn more about dissociation here:

brainly.com/question/25854432

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2 years ago

How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

Ksenya-84 [330]

Answer:

a. Approximately $1.3\; \rm mL$ .

b. Approximately $7.2\; \rm mL$ .

Explanation:

The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

Solution in a.:

$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

8 0

3 years ago