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3 years ago

Which of these elements has the highest first ionization energy?

Chemistry

1 answer:

julia-pushkina [17]3 years ago

6 0

Answer:

Cesium (Cs)

Explanation:

This eliment has an ionization energy of 3,8939

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Convert 300c into kelvin and Fahrenheit scale

Nitella [24]

Explanation:

300c = 1.8f+32= please calculate it yourself ☺

300c= k+273= 573k

5 0

3 years ago

A calorimeter contains 280.0g of water at 13.5C. When 0.250g of a substance with a molar mass of 510.0g/mol is dissolved, the te

marin [14]

The change in heat is simply equal to:

change in heat ΔH = final enthalpy – initial enthalpy

ΔH = [280.25 g * 4.18J/gC * (17.5°C)] – [280 g * 4.18J/gC * 13.5°C]

ΔH = 4,699.89 J = 4.7 kJ

<span>Hence heat released is about 4.7 kJ</span>

8 0

3 years ago

Name of 2 elements with their symbols derived from its german name

11Alexandr11 [23.1K]

Answer: Sodium (Na – Natrium)

Potassium (K – Kalium)

7 0

3 years ago

Harvey kept a balloon with a volume of 348 milliliters at 25.0˚C inside a freezer for a night. When he took it out, its new volu

ad-work [718]

Initial volume of the balloon = $V_{1}$ = 348 mL

Initial temperature of the balloon $T_{1}$ = $25.0^{0}C + 273 = 298 K$

Final volume of the balloon $V_{2}$ = 322 mL

Final temperature of the balloon = $T_{2} = ?$

According to Charles law, volume of an ideal gas is directly proportional to the temperature at constant pressure.

$\frac{V_{1} }{T_{1} } =\frac{V_{2} }{T_{2} }$

On plugging in the values,

$\frac{348mL}{298 K} =\frac{322 mL}{T_{2} }$

$T_{2} =276 K$

Therefore, the temperature of the freezer is 276 K

5 0

3 years ago

CH4 + 202 → CO2 + 2H2O<br> How many grams of O2 needed to produce 36 grams of H2O?

Kipish [7]

<h3>Answer:</h3>

64 g O₂

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<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>

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 36 g H₂O

[Solve] x g O₂

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

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

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

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

Set up conversion: $\displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \ H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})$
Divide/Multiply [Cancel Units]: $\displaystyle 63.929 \ g \ O_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

63.929 g O₂ ≈ 64 g O₂

8 0

3 years ago

Read 2 more answers