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trasher [3.6K]
3 years ago
9

Potassium carbonate, K 2CO 3, sodium iodide, NaI, potassium bromide, KBr, methanol, CH 3OH, and ammonium chloride, NH 4Cl, are s

oluble in water. Which produces the largest number of dissolved particles per mole of dissolved solute
Chemistry
1 answer:
slava [35]3 years ago
3 0

Answer:

Potassium carbonate (K₂CO₃)

Explanation:

The compounds dissociate into ions in water, as follows:

K₂CO₃ → 2 K⁺ + CO₃⁻    ⇒ 3 dissolved particles per mole

NaI → Na⁺ + I⁻    ⇒ 2 dissolved particles per mole

KBr → K⁺ + Br⁻   ⇒ 2 dissolved particles per mole

CH₃OH → CH₃O⁻ + H⁺  ⇒ 2 dissolved particles per mole

NH₄Cl → NH₄⁺ + Cl⁻   ⇒ 2 dissolved particles per mole

Therefore, the largest number of dissolved particles per mole of dissolved solute is produced by potassium carbonate (K₂CO₃).

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Magnesium hydroxide (Mg(OH)2): g/mol
Eva8 [605]

To determine the molar mass, you need to get the atomic mass of the molecule. To do this, check the periodic table for the atomic mass or average atomic weight of each element.

Mg = 24.305 x 1 = 24.305 amu

O = 15.9994 x 2 =31.9988 amu

H = 1.0079 x 2 = 2.0158 amu

 

Then, add all the components to get the atomic mass of the molecule.

24.305 amu + 31.9988 amu + 2.0158 amu = 58.3196 amu


The atomic mass is just equivalent to its molar mass.

So, the molar mass of Magnesium hydroxide (Mg(OH)2) is 58.3196 g/mol.

8 0
3 years ago
Read 2 more answers
Humans have the ability to change characteristics,or traits,of organism over time. What is this process called?
Vladimir79 [104]

Answer:

Natural Selection

7 0
3 years ago
What is the mass of 1.58 moles of CH4
HACTEHA [7]
<h3>Answer:</h3>

25.4 g CH₄

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.58 mol CH₄

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

Molar Mass of CH₄ - 12.01 + 4(1.01) = 16.05 g/mol

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 1.58 \ mol \ CH_4(\frac{16.05 \ g \ CH_4}{1 \ mol \ CH_4})
  2. Multiply/Divide:                 \displaystyle 25.359 \ g \ CH_4

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

25.359 g CH₄ ≈ 25.4 g CH₄

4 0
3 years ago
Assume you mix 100.0 mL of 200 M CsOH with 50.0 mL 0f 0.400 M HCl in a coffee cup calorimeter. A reaction occurs. The temperatur
Varvara68 [4.7K]

Answer:

The enthalpy change for the reaction per mole of CsOH is -56.1 kJ/mol

Explanation:

Step 1: Data given

Volume of a CsOH solution = 100.0 mL

Molarity of a CsOH solution = 0.200 M

Volume of HCl solution = 50.0 mL

Molarity of HCl solution = 0.400M

The temperature of both solutions before mixing was 22.50 °C, and it rises to 24.28 °C after the acid-base reaction.

Density = 1.00 g/mL

Specific heat = 4.2 J/gK = 4.2 J/g°C

Step 2: The balanced equation

CsOH + HCl → CsCl + H2O

Step 3: Calculate the energy

Q = m*c*ΔT

⇒with Q = the heat transfer = TO BE DETERMINED

⇒with m = the mass of the solution = (100+ 50 mL) * 1.00 g/mL = 150 grams

⇒with c = the specific heat of the solution = 4.2 J/g°C

⇒with ΔT = The change of temperature = T2 - T1  = 24.28 °C - 22.50 °C = 1.78 °C

Q = 150 grams * 4.2 J/g°C * 1.78 °C

Q = 1121.4 J

Step 4: Calculate moles CsOH

Moles CsOH = molarity CsOH * volume CsOH

Moles CsOH = 0.200M * 0.100 L

Moles CsOH = 0.0200 moles

Step 5: Calculate  the enthalpy change for the reaction per mole of CsOH

ΔH is negative since this is an exothermic reaction

ΔH = -Q/moles

ΔH = -1121.4 J / 0.0200 moles

ΔH = -56070 J/mol = -56.1 kJ/mol

The enthalpy change for the reaction per mole of CsOH is -56.1 kJ/mol

4 0
3 years ago
SO
Burka [1]

Answer:

\large \boxed{\text{E) 721 K; B) 86.7 g}}

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL;  V₂ = 435 mL

T₁ = 355 K;     T₂ = ?

b) Calculation

\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}

7 0
3 years ago
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