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ANEK [815]
2 years ago
10

PLEASE HELP WILL GIVE BRAINLIEST

Chemistry
2 answers:
Phoenix [80]2 years ago
5 0

Answer:

Coefficients represents no of moles while subscripts represent no of atoms.

stepan [7]2 years ago
5 0

Answer:

Coefficients came before formula . Example = 4CO means there are 4 molecules of carbonmonooxide.

Subscripts came after and below the formula . Example = Oxygen gas in which 2 is a subscript which means there are 2 atoms of Oxygen in 1 molecule of Oxygen i.e. oxygen gas

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If a gas occupies 79.5 mL at -1.4°C, what temperature, in Kelvin, would it
Anna35 [415]

Answer:

121 K

Explanation:

Step 1: Given data

  • Initial volume (V₁): 79.5 mL
  • Initial temperature (T₁): -1.4°C
  • Final volume (V₂): 35.3 mL

Step 2: Convert "-1.4°C" to Kelvin

We will use the following expression.

K = °C + 273.15 = -1.4°C + 273.15 = 271.8 K

Step 3: Calculate the final temperature of the gas (T₂)

Assuming ideal behavior and constant pressure, we can calculate the final temperature of the gas using Charles' law.

V₁/T₁ = V₂/T₂

T₂ = V₂ × T₁/V₁

T₂ = 35.3 mL × 271.8 K/79.5 mL = 121 K

5 0
3 years ago
When salt is dissolved in water, the salt particles are referred to as the _____.
Sergio039 [100]
The salt is a solid compound and is considered the "Solute" of the solution.
5 0
3 years ago
Read 2 more answers
These biological compounds are nonpolar and insoluble in water:
juin [17]
The biological compounds that are nonpolar and insoluble in water are lipids. It is a group of molecules that are naturally occurring which includes sterols, waxes, fats, fat-soluble vitamins and the like. These molecules are nonpolar molecules so basically the cannot be dissolved in a polar solvent like water.
6 0
3 years ago
Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co
amid [387]

Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as p_{d} and the proportion of air that is by-passed as p_{bp}, being p_{d}+p_{bp}=1.

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

0.004 = 0.016*p_{bp}+ 0.002*p_{d}

Replacing the first equation in the second one we have

0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.

The water removed of every 100 kg of dry air is

100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

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
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