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3 years ago

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All compounds are molecules but not all molecules are compounds. Explain this statement.

1 answer:

Elis [28]3 years ago

5 0

All compounds are molecules because a molecule is 2 or more substances/elements combined and a compound is 2 or more elements combined. But not all molecules are elements because some molecules are just combined substances with no elements combined at all.

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4 years ago

The gas inside a balloon exerts a pressure of 3atm at 298 K. What is the temperature required to increase the pressure to 6 atm?

podryga [215]

Answer: 596 atm

Explanation:

Given that,

Original pressure of balloon P1 = 3atm

Original temperature of balloon T1 = 298K

New pressure of balloon P2 = 6atm

New temperature of balloon T2 = ?

Since pressure and temperature are given while volume is constant, apply the formula for Pressure law

P1/T1 = P2/T2

3 atm / 298K = 6 atm / T2

To get the value of T2, cross multiply

3 atm x T2 = 6 atm x 298K

3 atmT2 = 1788 atmK

Divide both sides by 3 atm

3 atmT2 / 3 atm = 1788 atmK / 3 atm

T2 = 596 atm

Thus, a temperature of 596 atmospheres is required to increase the pressure to 6 atm.

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3 years ago

It takes a person 35 seconds to swim in a straight line from the east end of a pool to the west end of the pool, a distance of 5

viktelen [127]

Answer:

1.42m/s

Explanation:

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3 years ago

Different types of plants use different types of carbon fixation as adaptations to varying conditions.

monitta

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3 years ago

Read 2 more answers

A chemist titrates 120.0 mL of a 0.4006 M hydrocyanic acid (HCN) solution with 0.6812 MNaOH solution at 25 °C. Calculate the pH

lukranit [14]

Explanation:

The given reaction equation is as follows.

$HCN + NaOH \rightarrow NaCN + H_{2}O$

Hence, initial moles of HCN will be as follows.

Moles = Molarity × Volume

= $0.4006 M \times 120 ml$

= 48.072 mol

Now, we will calculate the volume of NaOH as follows.

$M_{1}V_{1} = M_{2}V_{2}$

$0.4006 \times 120 = 0.6812 \times V_{2}$

$V_{2}$ = 70.57 ml

At the equivalence point, moles of both HCN and NaOH will be equal. And, total volume will be as follows.

120 ml + 70.57 ml = 190.57 ml

Initial concentration of $CN^{-}$ is as follows.

$\frac{48.072}{190.57}$

= 0.25 M

Now, the equilibrium equation will be as follows.

$CN^{-} + H_{2}O \rightleftharpoons HCN + OH^{-}$

Initial: 0.25

Equilbm: 0.25 - x x x

Expression to find $K_{a}$ is as follows.

$K_{a} = \frac{x^{2}}{0.25 - x}$

We know that, $pK_{a} = -log K_{a}$

$K_{a}$ = antilog (-9.21)

= $6.16 \times 10^{-10}$

As, $K_{b} = \frac{K_{w}}{K_{a}}$

$\frac{10^{-14}}{6.16 \times 10^{-10}} = \frac{x^{2}}{0.25 - x}$

x = $0.2 \times 10^{-2} = [OH^{-}]$

Now, we will calculate the pH as follows.

pOH = $-log[OH^{-}]$

as pOH = 2.698

And, pH + pOH = 14

pH = 14 - 2.698

= 11.30

Thus, we can conclude that pH at equivalence is 11.30.

4 0

3 years ago