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Will the boiling point of water be higher or lower in the maintains, where the altitude is high, and the pressure is low ?

Vlad [161]

Higher low preccure

4 0

2 years ago

using the soubility curve what is the solubilityof nh4cl in 10 mL of water at a temperature of 60 degrees Celsius

lukranit [14]

Answer:

Please, see attached two figures:

The first figure shows the solutility curves for several soluts in water, which is needed to answer the question.

The second figure shows the reading of the solutiblity of NH₄Cl at a temperature of 60°C.

Answer: 5.5g

Explanation:

The red arrow on the second attachement shows how you must go vertically from the temperature of 60ºC on the horizontal axis, up to intersecting curve for the solubility of NH₄Cl.

From there, you must move horizontally to the left (green arrow) to reach the vertical axis and read the solubility: the reading is about in the middle of the marks for 50 and 60 grams of solute per 100 grams of water: that is 55 grams of grams of solute per 100 grams of water.

Assuming density 1.0 g/mol for water, 10 mL of water is:

$10mL\times 1.0g/mL=10g$

Thus, the solutibily is:

$10gWater\times 55gNH_4Cl/100gWater=5.5gNH_4Cl$

5 0

2 years ago

in example 5.11 of the text the molar volume of n2 at STP is given as 22.42 L/mol how is this number calculatd how does the mola

Valentin [98]

Answer:

V = 22.42 L/mol

N₂ and H₂ Same molar Volume at STP

Explanation:

Data Given:

molar volume of N₂ at STP = 22.42 L/mol

Calculation of molar volume of N₂ at STP = ?

Comparison of molar volume of H₂ and N₂ = ?

Solution:

Molar Volume of Gas:

The volume occupied by 1 mole of any gas at standard temperature and pressure and it is always equal to 22.42 L/ mol

Molar volume can be calculated by using ideal gas formula

PV = nRT

Rearrange the equation for Volume

V = nRT / P . . . . . . . . . (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

Standard values

P = 1 atm

T = 273 K

n = 1 mole

R = 0.08206 L.atm / mol. K

Now put the value in formula (1) to calculate volume for 1 mole of N₂

V = 1 x 273 K x 0.08206 L.atm / mol. K / 1 atm

V = 22.42 L/mol

Now if we look for the above calculation it will be the same for H₂ or any gas. so if we compare the molar volume of 1 mole N₂ and H₂ it will be the same at STP.

6 0

2 years ago

. How many milliliters of 0.20 M HCl are needed to exactly neutralize 40. milliliters of 0.40 M KOH

Anuta_ua [19.1K]

Answer:

$V_{HCl}=80mL$

Explanation:

Hello,

In this case, for the given reactants we identify the following chemical reaction:

$KOH+HCl\rightarrow KCl+H_2O$

Thus, we evidence a 1:1 molar ratio between KOH and HCl, therefore, for the complete neutralization we have equal number of moles, that in terms of molarities and volumes become:

$n_{HCl}=n_{KOH}\\\\M_{HCl}V_{HCl}=M_{KOH}V_{KOH}$

Hence, we compute the volume of HCl as shown below:

$V_{HCl}=\frac{M_{KOH}V_{KOH}}{M_{HCl}} =\frac{0.40M*40mL}{0.20M} \\\\V_{HCl}=80mL$

Best regards.

5 0

2 years ago

How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 1

valentinak56 [21]

4 Al + 3 O2 → 2 Al2O3

(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2

0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.

((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over

7 0

3 years ago

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