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djverab [1.8K]
3 years ago
10

If 1.2 moles of ideal gas occupy a volume of 18.2 L at a pressure of 1.8 ATM what is the temperature of gas in degrees Celsius.

Chemistry
1 answer:
Novosadov [1.4K]3 years ago
8 0

Answer: A 59.5 degree celcius

The equation that we will use to solve this problem is :

PV = nRT where:

P is the pressure of gas = 1.8 atm

V is the volume of gas = 18.2 liters

n is the number of moles of gas = 1.2 moles

R is the gas constant =  0.0821

T is the temperature required (calculated in kelvin)

Using these values to substitute in the equation, we find that:

(1.8)(18.2) = (1.2)(0.0821)(T)

T = 332.5 degree kelvin

The last step is to convert the degree kelvin into degree celcius:

T = 332.5 - 273 = 59.5 degree celcius

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If 20 grams of Zinc phosphate reacts with excess hydrochloric acid and produces 18 grams of Zinc chloride what is the percent yi
fenix001 [56]

Answer:

Y=85\%

Explanation:

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In this case, since we know the balanced chemical reaction, we are first able to realize there is a 1:3 mole ratio between zinc phosphate and zinc chloride; it means that we can first compute the moles of the desired product via stoichiometry:

n_{ZnCl_2}=20gZn_3(PO_4)_2*\frac{1molZn_3(PO_4)_2}{386.11gZn_3(PO_4)_2}*\frac{3molZnCl_2}{1molZn_3(PO_4)_2}=0.16gZnCl_2

Next, since those moles are associated with the theoretical yield of zinc chloride, we obtain the corresponding mass:

m_{ZnCl_2}^{theoretical}=0.16molZnCl_2*\frac{136.29gZnCl_2}{1molZnCl_2} =21gZnCl_2

Finally, we compute the percent yield by diving the actual yield (18 g) by the theoretical yield:

Y=\frac{18g}{21g}*100\%\\\\Y=85\%

Best regards!

4 0
3 years ago
HELLPPP PLZZ ASAPP!
Sphinxa [80]

Answer:

Condensation and Depositon

Explanation:

Condensation is from gas to liquid

Deposition is from gas to solid

7 0
3 years ago
Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first i
Law Incorporation [45]

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹;    5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.  

                     AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹:                               s               s

K_{sp} =\text{[Ag$^{+}$][IO$_{3}$$^{-}$]} = s\times s =  s^{2} = 3.0\times 10^{-8}\\s = \sqrt{3.0\times 10^{-8}} \text{ mol/L} = 1.7 \times 10^{-4} \text{ mol/L}

b. Barium sulfate

                     BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹:                                0.02             0

C/mol·L⁻¹:                                 +s              +s

E/mol·L⁻¹:                            0.02 + s          s

K_{sp} =\text{[Ba$^{2+}$][SO$_{4}$$^{2-}$]} = (0.02 + s) \times s \approx  0.02s = 1.1\times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{0.02} \text{ mol/L} = 5.5 \times 10^{-9} \text{ mol/L}

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is  

\mu = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\\mu = \dfrac{1}{2} (\text{[Ba$^{2+}$]}\cdot (2+)^{2} + \text{[NO$_{3}$$^{-}$]}\times(-1)^{2}) = \dfrac{1}{2} (\text{0.02}\times 4 + \text{0.04}\times1)= \dfrac{1}{2} (0.08 + 0.04)\\\\= \dfrac{1}{2} \times0.12 = 0.06

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(1)^{2}\sqrt{0.06} = -0.51\times 0.24 = -0.12\\\gamma = 10^{-0.12} = 0.75

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

K_{sp} =\text{[Ag$^{+}$]$\gamma_{Ag^{+}}$[IO$_{3}$$^{-}$]$\gamma_{IO_{3}^{-}}$} = s\times0.75\times s \times 0.75 =0.56s^{2}= 3.0 \times 10^{-8}\\s^{2} = \dfrac{3.0 \times 10^{-8}}{0.56} = 5.3 \times 10^{-8}\\\\s =2.3 \times 10^{-4}\text{ mol/L}

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(2)^{2}\sqrt{0.06} = -0.51\times16\times 0.24 = -0.50\\\gamma = 10^{-0.50} = 0.32

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

K_{sp} =\text{[Ba$^{2+}$]$\gamma_{ Ba^{2+}}$[SO$_{4}$$^{2-}$]$\gamma_{ SO_{4}^{2-}}$} = (0.02 + s) \times 0.32\times s\times 0.32 \approx  0.02\times0.10s\\2.0\times 10^{-3}s = 1.1 \times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{2.0 \times 10^{-3}} \text{ mol/L} = 5.5 \times 10^{-8} \text{ mol/L}

7 0
3 years ago
Which characteristic do valence electrons indicate about reactions between atoms?
AURORKA [14]

Answer:

the answer is Bond Number

Explanation:

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6 0
3 years ago
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How many coulombs are required to plate a layer of chromium metal 0.28 mm thick on an auto bumper with a total area of 0.40 m2 f
Rufina [12.5K]

Explanation:

The given data is as follows.

Thickness = 0.28 mm = 0.28 \times \frac{1}{10} cm = 0.028 cm

Area = 0.40 m^{2} = 0.40 \times 10^{4} cm^{2} = 4000 cm^{2}

As, it is known that volume = area × thickness

So,             Volume = 4000 cm^{2} \times 0.028 cm      

                                = 112 cm^{3}

As density is mass divided by volume. So, mass of chromium will be calculated as follows.

                   Density = \frac{mass}{volume}  

            7.20 g/cm^{3} = \frac{mass}{112 cm^{3}}      

                      mass = 806.4 g

As, mass of 1 mole of chromium is 52 g. So, number of moles in 806.4 g of chromium will be as follows.

              No. of moles = \frac{mass}{molar mass}                            

                                     = \frac{806.4 g}{52 g}

                                     = 15.50 mol

In chromate ion, (CrO^{2-}_{4}) charge on Cr is +6. It means that 6 electrons are needed to reduce Cr^{+6} into Cr.

As, 1 mole of Cr^{+6} ions require 6 moles of electrons. Therefore, moles of electrons for 15.50 mol will be calculated as follows.

                              6 × 15.50 mol = 93.04 mol

To calculate number of electrons we multiply number of moles by Avogadro's number as follows.

               93.04 mol \times 6.02 \times 10^{23}

                      = 560.13 \times 10^{23}

                     = 5.6 \times 10^{25} electrons

There is magnitude of 6.241 \times 10^{18} times the charge on an electron is equal to 1 coulomb.

Hence, number of coulombs will be as follows.

              No. of coulombs = \frac{5.6 \times 10^{25}}{6.241 \times 10^{18}}

                                           = 0.897 \times 10^{7} C

or,                                        = 8.97 \times 10^{6} C

Thus, we can conclude that 8.97 \times 10^{6} C are required to plate a layer of chromium metal with given data.

4 0
2 years ago
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