All categories

2 years ago

Convert a pressure of 0.0248 mm Hg to the equivalent pressure in pascals (Pa).

Chemistry

1 answer:

drek231 [11]2 years ago

5 0

Answer is 3.306 Pa.

<em>Explanation:</em>

This is a simple unit conversion problem.

1 atm = 760 mmHg = 101325 Pa

Hence,

0.0248 mmHg = (101325 Pa / 760 mmHg) x 0.0248 mmHg

= 3.306 Pa

Pressure can be measured by using many units such as atm, Pa, torr, bar, cmHg, mmHg and so on.

The relationships among those units as follows

1 atm = 101325 Pa = 1.01325 bar = 760 torr, 760 mmHg = 76 cmHg

You might be interested in

Question 1

Nezavi [6.7K]

Answer:

The heat of reaction is $Q = 170.92 \ J$

The enthalpy is $\Delta H = -170.91 \ J$

The concentration of HCl is $M__{HCl}} =2.468 M$

Explanation:

From the question we are told that

The volume of $NaOH$ is $V_{NaOH} = 45 \ cm^3 = \frac{45} {1000} = 0.045 L$

The number of concentration of $NaOH$ is $M__{NaOH}} = 2M$

The volume of HCl is $V_{HCl} = 20 cm^3$

The number of concentration of $HCl$ is $M__{HCl}} = 1 M$

The temperature difference is $\Delta T = 9.4 ^o C$

Now the heat of reaction is mathematically represented as

$Q = m * c_p * \Delta T$

Where

$c_p$ is the specific heat of water with value $c_p = 4200 J /kg ^o C$

$m = m__{NaOH}} + m__{HCl}}$

Now $m__{NaOH}} = V_{NaOH} * M_{NaOH} * Z_{NaOH}$

where $Z_{NaOH}$ is the molar mass of NaOH with the value of 0.04 kg/mol

So $m__{NaOH}} = 0.045 * 2 * 0.040$

$m__{NaOH}} =0.0036\ kg$

While

$m__{HCl}} = V_{HCl} * M_{HCl} + Z_{HCl}$

Where $Z_{HCl}$ is the molar mass of $HCl$ with the value of 0.03646 kg/mol

$m__{NaOH}} = 0.020 * 1 * 0.03646$

$m__{NaOH}} = 0.000729 kg$

$m = 0.0036 + 0.000729$

$m = 0.00433$

=> $Q = 0.00433 * 4200 * 9.4$

$Q = 170.92 \ J$

The enthalpy is mathematically represented as

$\Delta H = - Q$

=> $\Delta H = -170.91 \ J$

From the second question we are told that

The volume of HCl is $V_{HCl_1} = 10cm^3 = \frac{10}{1000} = 0.010 L$

The volume of NaOH is $V_{NaOH_1 } = 30 cm^3 = \frac{30}{1000} = 0.03 L$

The concentration of NaOH is $M_{NaOH} = 2 M$

The first temperature change is $\Delta T = 4.5 ^oC$

The second volume of $V_{HCl_2} = 20 cm^3 = \frac{20}{1000 } = 0.020 m^3$

The mass of NaOH is

$m__{NaOH}} = V_{NaOH} * M_{NaOH} * Z_{NaOH}$

substituting values

$m__{NaOH}} = 0.03 * 2 * 40$

$m__{NaOH}} = 3.6 \ g$

The mass of the product formed is

$m = \frac{Q}{c_p * \Delta T}$

substituting values

$m = \frac{170.91}{4200 * 9} * 1000$

The multiplication by 1000 is to convert it from kg to grams

$m = 4.5 g$

Now the mass of HCl is

$m__{HCl}} = m - m__{NaOh}}$

substituting values

$m__{HCl}} = 4.5 -3.6$

$m__{HCl}} = 0.9 \ g$

Now the concentration of HCl is

$M__{HCl}} = \frac{m_{HCl}}{(Z_{HCl} * *1000) * V_{HCl_1}}}$

The multiplication of $Z_{HCl}$ is to convert it from kg/mol to g/mol

$M__{HCl}} = \frac{0.9}{36.46 * 0.01}}$

$M__{HCl}} =2.468 M$

3 0

2 years ago

The acid-dissociation constant, ka, for gallic acid is 4.57 ⋅ 10-3. what is the base-dissociation constant, kb, for the gallate

LiRa [457]

Hello!

To determine the Kb of gallic acid is actually very simple.

The dissociation reaction of Gallic Acid (HGal) is the following:

HGal+H₂O ⇄ H₃O⁺ + Gal⁻

The equation for converting from Ka to Kb is the following:

$Ka*Kb=Kw \\ \\ Kb= \frac{Kw}{Ka}= \frac{1*10^{-14} }{4,57*10^{-3} }=2,19 * 10^{-12}$

So, the Kb is 2,19*10⁻¹²

Have a nice day!

3 0

2 years ago

Read 2 more answers

How do genetic factors influence the formation of "dead zones” around creosote bushes?

Korvikt [17]

Answer:

Genetic factors contributes to the formation of the dead zones is by their ability to expand their abilities that enables them to spread and contribute to the formation of the dead zones. These dead zones are made when the oxygen are low where it is necessarily important for the aquatic life, if the oxygen needed is depleted or too low, instead of supporting aquatic life, dead zones are created and factors contribute to these occurrences with their ability to expand.

Explanation:

5 0

2 years ago

Read 2 more answers

Consider the following chemical reaction: CO (g) + 2H2(g) ↔ CH3OH(g) At equilibrium in a particular experiment, the concentratio

AlexFokin [52]

Answer:

The equilibrium concentration of CH₃OH is 0.28 M

Explanation:

For the reaction: CO (g) + 2H₂(g) ↔ CH₃OH(g)

The equilibrium constant (Keq) is given for the following expresion:

Keq= $\frac{(CH3OH)}{(CO) x (H2)^{2}}$ =14.5