Why is it important to know what temperature scale is being used in a given situation?

Answers

Answer 1

Answer: It depends, for example, it is quite important to know the Kelvin scale (i.e 0 degrees Celsius is 273 K and -273 degrees Celsius is 0 K ) when dealing gases. But I don't know other situations where you would need to know other temperature scales.

Hope this helps and also if you are using Fahrenheit 1 Fahrenheit is -17.22 degrees Celsius

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If you mix different amounts of two ideal gases that are originally at different temperatures, what must be true of the final state after the temperature stabilizes? (There may be more than one correct choice.) a) Both gases will reach the same final temperature.

b) The final rms molecular speed will be the same for both gases.

c) The final average kinetic energy of a molecule will be the same for both gases.

Answers

Answer:

a,c are correct

Explanation:

a) On mixing two gases the final temperature of both the gases becomes the same. The heat will flow from high temp. gas to lower temp gas till the temp of both gases become equal (Thermal equilibrium). This is correct.

b) The rms speed of the molecule is inversely proportional to its molar mass so the final rsm will not be the same. This is incorrect.

c) The average kinetic energy of the system will remain the same. Hence this is also correct.

What is the empirical formula of a compound that is 64.3 % c, 7.2 % h, and 28.5 % o by mass?

Answers

Asnwer : Empirical formula of a compound is : $C_(3)H_(4)O$

Given information : C = 64.3 % , H = 7.2 % , O = 28.5 %

Step 1 : Convert the given percentage (%) to grams.

Explanation : Let the total mass of the compound be 100 grams.

Mass of C = 64.3 g

$(100g)* ((64.3percent))/((100percent)) = 64.3g$

Mass of H = 7.2 g

$(100g)* ((7.2percent))/((100percent)) = 7.2g$

Mass of O = 28.5 g

$(100g)* ((28.5percent))/((100percent)) = 28.5g$

Step 2 : Convert the grams of each compound to moles.

$Moles = (Grams)/(Molar mass)$

Molar mass of C = 12.0g/mol

Molar mass of H = 1.0 g/mol

Molar mass of O = 16.0g/mol

$Moles of C = (64.3g)/(12.0(g)/(mol))$

Moles of C = 5.36 mol

$Moles of H = (7.2g)/(1.0(g)/(mol))$

Moles of H = 7.2 mol

$Moles of O = (28.5g)/(16.0(g)/(mol))$

Moles of O = 1.78 mol

Step 3 : Find the mole ratio of C , H and O

Mole ratio is calculated by dividing the mole values by the smallest value.

Mole of C = 5.36 mol , Mole of H = 7.2 mol , Mol of O = 1.78 mol

Out of the three mole values , mole value of O that is 1.78 mol is less , so we divide all the mole values by 1.78 mol.

$Mole of C = (5.36mol)/(1.78mol) = 3.0$

$Mole of H = (7.2mol)/(1.78mol) = 4.0$

$Mole of O = (1.78mol)/(1.78mol) = 1.0$

C : H : O = 3 : 4 : 1

So empirical formula of the compound is $C_(3)H_(4)O_(1)$ or $C_(3)H_(4)O$

Determine the concentrations of K2SO4, K+, and SO42− in a solution prepared by dissolving 2.07 × 10−4 g K2SO4 in 2.50 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per million (ppm). Note: Determine the formal concentration of SO42−. Ignore any reactions with water.

Answers

Answer:

[K2SO4] = 4,75x10⁻⁷M ; [K⁺] = 9.50x10⁻⁷M ; [SO4⁻²] = 4,75x10⁻⁷M

SO4⁻²: 0.045ppm ; K⁺: 0.037ppm

[SO4⁻²] = 4,70x10⁻⁷ F

Explanation:

Determine the equation

K2SO4 → 2K⁺ + SO4⁻²

Each mole of potassium sulfate generates two moles of potassium cation and one mole of sulfate anion

Molar mass K2SO4: 174.26 g/m

Moles of K2SO4: grams / molar mass

2.07x10⁻⁴g / 174.26 g/m = 1.18x10⁻⁶ moles

Molarity: Moles of solute in 1 L of solution

1.18x10⁻⁶ moles / 2.5 L = 4,75x10⁻⁷M (K2SO4)

K⁺ : 4,75x10⁻⁷M . 2 = 9.50x10⁻⁷M

SO4⁻²: 4,75x10⁻⁷ M

1 mol of K2SO4 has 2 moles of K and 1 mol of SO4

1.18x10⁻⁶ moles of K2SO4 has 1.18x10⁻⁶ moles of SO4 and 2.37x10⁻⁶ moles of K.

1.18x10⁻⁶ moles of SO4⁻² are 1.13x10⁻⁴ grams (moles. molar mass)

2.37x10⁻⁶ moles of K are 9.26x10⁻⁵ grams (moles. molar mass)

These grams are in 2.5 L of water, so we need μg/mL to get ppm

2.5 L = 2500 mL

1.13x10⁻⁴ grams SO4⁻² are 113.35 μg (1 μg = 1x10⁶ g)

9.26x10⁻⁵ grams K⁺ are 92.6 μg (1 μg = 1x10⁶ g)

113.35 μg /2500 mL = 0.045ppm

92.6 μg /2500 mL = 0.037ppm

Formal concentration of SO4⁻² :

Formality = Number of formula weight of solute / Volume of solution (L)

(1.13x10⁻⁴ grams / 96.06 g ) / 2.5 L = 4,70x10⁻⁷ F

a length of #8 copper wire (radius = 1.63 mm) has a mass of 24.0 kg and a resistance of 2.061 ohm per km. what is the overall resistance of the wire

Answers

m = pVol

p : copper density

Vol: volume

Vol = Al

A: area

L: length

m = p * A * L

L = m/p*A

To find the total resistance

Rtotal = R per Km * L/1000

Actually I think the question is missing the copper density.

How many valence electrons must two atoms share to form a single covalent bond? answers A.2 B.4 C.3 D.1

Answers

Answer:

Explanation:

A single covalent bond is formed when two electrons are shared between the same two atoms, one electron from each atom.

Answer:

the answer is 2

Explanation:

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first ignoring ionic strength and activities. a. silver iodate
b. barium sulfate
c. Repeat the above calculations using ionic strength and activities.

Answers

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)$$^(-)$]} = s* s = s^(2) = 3.0* 10^(-8)\ns = \sqrt{3.0* 10^(-8)} \text{ mol/L} = 1.7 * 10^(-4) \text{ mol/L}$

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-)$]} = (0.02 + s) * s \approx 0.02s = 1.1* 10^(-10)\ns = (1.1* 10^(-10))/(0.02) \text{ mol/L} = 5.5 * 10^(-9) \text{ mol/L}$

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

$\mu = (1)/(2) \sum_(i) {c_(i)z_(i)^(2)}\n\n\mu = (1)/(2) (\text{[Ba$^(2+)$]}\cdot (2+)^(2) + \text{[NO$_(3)$$^(-)$]}*(-1)^(2)) = (1)/(2) (\text{0.02}* 4 + \text{0.04}*1)= (1)/(2) (0.08 + 0.04)\n\n= (1)/(2) *0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(1)^(2)√(0.06) = -0.51* 0.24 = -0.12\n\gamma = 10^(-0.12) = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_(sp) =\text{[Ag$^(+)$]$\gamma_(Ag^(+))$[IO$_(3)$$^(-)$]$\gamma_{IO_(3)^(-)}$} = s*0.75* s * 0.75 =0.56s^(2)= 3.0 * 10^(-8)\ns^(2) = (3.0 * 10^(-8))/(0.56) = 5.3 * 10^(-8)\n\ns =2.3 * 10^(-4)\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(2)^(2)√(0.06) = -0.51*16* 0.24 = -0.50\n\gamma = 10^(-0.50) = 0.32$

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-)}$} = (0.02 + s) * 0.32* s* 0.32 \approx 0.02*0.10s\n2.0* 10^(-3)s = 1.1 * 10^(-10)\ns = (1.1* 10^(-10))/(2.0 * 10^(-3)) \text{ mol/L} = 5.5 * 10^(-8) \text{ mol/L}$

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