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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.

Explanation:

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

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2.5 moles of sodium chloride is dissolved to make 0.050 liters of solution.

ivanzaharov [21]

The answer is:

the molarity = 50 moles/liters

The explanation:

when the molarity is = the number of moles / volume per liters.

and when the number of moles =2.5 moles

and the volume per liters = 0.05 L

so by substitution:

the molarity = 2.5moles/0.05L

= 50 moles /L

MARK ME BRAINLIEST PLEASE!!!!!!!!!

3 0

2 years ago

How many grams of antifreeze C2H4(OH)2 would be required per 500 g of water to prevent the water from freezing at a temperature

Wewaii [24]

Answer:

333.7 g.

Explanation:

The depression in freezing point of water (ΔTf) due to adding a solute to it is given by: ΔTf = Kf.m.

Where, ΔTf is the depression in water freezing point (ΔTf = 20.0°C).

Kf is the molal freezing point depression constant of the solvent (Kf = 1.86 °C/m).

m is the molality of the solution.

∴ m = ΔTf/Kf = (20.0°C)/(1.86 °C/m) = 10.75 m.

molaity (m) is the no. of moles of solute per kg of the solvent.

∵ m = (no. of moles of antifreeze C₂H₄(OH)₂)/(mass of water (kg))

∴ no. of moles of antifreeze C₂H₄(OH)₂ = (m)(mass of water (kg)) = (10.75 m)(0.5 kg) = 5.376 mol.

∵ no. of moles = mass/molar mass.

∴ mass of antifreeze C₂H₄(OH)₂ = no. of moles x molar mass = (5.376 mol)(62.07 g/mol) = 333.7 g.

5 0

2 years ago

If 5.61 g of ammonium nitrate (NH4NO3) is dissolved in 155 mL of water to make an instant cold pack, what is the molarity of the

kifflom [539]

Answer:

0.452 M

Explanation:

Remember that to get molarity, you have to go thru an equation:

M=mol/V (in liters)

So...

5.61 g NH4NO3 x 1 mol NH4NO3/80.043 g

=0.0701 mol NH4NO3

Convert mL to L

=155--->0.155 L

Plug in and divide.

0.0701/0.155

=0.452 M

7 0

2 years ago

How many lead (Pb) atoms will be generated when 5.38 moles of ammonia react according to the following equation: 3PbO+2NH3→3Pb+N

jasenka [17]

Answer:

4.86×10^23 molecule of Pb

Explanation:

Based on that equation, for every 2 moles of ammonia, you get 3 moles of lead.

So:

2 mol NH3/ 3 mol Pb

Using this ratio we can find the amounts of either molecule. Given 5.38 mol NH3:

(5.38 NH3)(3 Pb/ 2 NH3) = (5.38)(3/2) mol Pb = 8.07 mol Pb

Then, we just need to use Avagadro's number to get the number of molecules.

(8.07)(6.02×10^23) = 4.86×10^23 molecule of Pb

4 0

2 years ago

Read 2 more answers

Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 m

Vesna [10]

Answer : The correct option is, (A) -101.37 KJ

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.1392 mole of HCl neutralizes by 0.1392 mole of NaOH

Thus, the number of neutralized moles = 0.1392 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $87ml+87ml=174ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 174ml=174g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 174 g

$T_{final}$ = final temperature of water = 317.4 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=174g\times 4.18J/g^oC\times (317.4-298)K$

$q=14110.008J=14.11KJ$

Thus, the heat released during the neutralization = -14.11 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -14.11 KJ

n = number of moles used in neutralization = 0.1392 mole

$\Delta H=\frac{-14.11KJ}{0.1392mole}=-101.37KJ/mole$

Therefore, the enthalpy of neutralization is, -101.37 KJ

3 0

2 years ago