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vazorg [7]
2 years ago
12

How much pressure would 0.8 moles of a gas at 370K exert if it occupied 17.3L of space

Chemistry
2 answers:
dezoksy [38]2 years ago
7 0

Answer:

1.40 atm is the pressure for the gas

Explanation:

An easy problem to solve with the Ideal Gases Law:

P . V = n . R .T

T° = 370K

V = 17.3L

n = 0.8 mol

Let's replace data → P . 17.3L = 0.8mol . 0.082L.atm/mol.K . 370K

P = (0.8mol . 0.082L.atm/mol.K . 370K) / 17.3L = 1.40 atm

DerKrebs [107]2 years ago
3 0

Answer:

The pressure is 1.40 atm

Explanation:

Step 1: Data given

Number of moles = 0.8 moles

Temperature = 370 K

volume = 17.3 L

Step 2: Calculate the pressure

p*V = n*R*T

⇒ with p = the pressure = TO BE DETERMINED

⇒with V = the volume = 17.3 L

⇒with n = the number of moles of gas = 0.8 moles

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 370 K

p = (n*R*T)/V

p = (0.8 * 0.08206 * 370) / 17.3 L

p = 1.40 atm

The pressure is 1.40 atm

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If you have 10cm of snow with a volume of 40mL and a density of 0.5 g/mL how many inches of rain is this?
Lelu [443]

Density is the mass of compound divided by its volume can be shown as follows:

d = 0.5 g /mL = m /V = m /40 = 0.5

m = 20 g

40 mL of snow having 20 g of mass calculated from density.

Now, 10 cm of snow = 3.93 inches = 20 g  

As, 10 inches of rain will produce 11 inches of ice as the volume of ice is bigger than rain water.

10 inches rain = 11 inches snow

3.93 inches of snow produced by  

= (3.93 * 10) /11 = 3.57 inches rain.

Thus, 3.57 incehs of rain produces by 10 cm snow.

4 0
2 years ago
Four animals were fleeing from a wildfire. All of them were running at a speed of 30 miles per hour. Which one has the greatest
RUDIKE [14]

Answer: I don’t know lol

Explanation:  I am so sorry I thought this was easy

6 0
3 years ago
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What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?​
UNO [17]

Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                      NH₃          :            N₂

                        4             :             2

                      0.12           :           2/4×0.12 = 0.06

                      NH₃         :            H₂O

                        4            :             6

                        0.12       :           6/4×0.12 = 0.18

                       

                       O₂            :            N₂

                        3             :             2

                      0.125        :           2/3×0.125 = 0.08

                        O₂           :            H₂O

                        3              :             6

                        0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                        NH₃          :             O₂

                           4            :              3

                           0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

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2 years ago
Calcium hydroxide is a strong base but is not very soluble ( Ksp = 5.02 X 10-6 ). What is the pH of a saturated solution of Ca(O
Misha Larkins [42]

Answer : The pH of a saturated solution is, 12.33

Explanation : Given,

K_{sp} = 5.02\times 10^{-6}

First we have to calculate the solubility of OH^- ion.

The balanced equilibrium reaction will be:

Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-

Let the solubility will be, 's'.

The concentration of Ca^{2+} ion = s

The concentration of OH^- ion = 2s

The expression for solubility constant for this reaction will be,

K_{sp}=[Ca^{2+}][OH^-]^2

Let the solubility will be, 's'

K_{sp}=(s)\times (2s)^2

K_{sp}=(4s)^3

Now put the value of K_[sp} in this expression, we get the solubility.

5.02\times 10^{-6}=(4s)^3

s=1.079\times 10^{-2}M

The concentration of Ca^{2+} ion = s = 1.079\times 10^{-2}M

The concentration of OH^- ion = 2s = 2\times (1.079\times 10^{-2}M)=2.158\times 10^{-2}M

First we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (2.158\times 10^{-2})

pOH=1.67

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1.67=12.33

Therefore, the pH of a saturated solution is, 12.33

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stira [4]

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