If energy is transferred from a table to a block of ice moving across the table, which of the following statements is true?

Which of the following is a molecular formula for a compound with an empirical formula of CH2O and a molar mass of 150. g/mol.

trapecia [35]

Answer:

C₅H₁₀O₅

Explanation:

Let's consider a compound with the empirical formula CH₂O. In order to determine the molecular formula, we have to calculate "n", so that

n = molar mass of the molecular formula / molar mass of the empirical formula

The molar mass of the molecular formula is 150 g/mol.

The molar mass of the empirical formula is 12 + 2 × 1 + 16 = 30 g/mol

n = (150 g/mol) / (30 g/mol) = 5

Then, we multiply the empirical formula by 5.

CH₂O × 5 = C₅H₁₀O₅

7 0

3 years ago

Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of a 0.100-M solution of acetic

Mademuasel [1]

Answer:

1.33%

Explanation:

In an aqueous solution, a weak acid such as acetic acid, will be in equilibrium with its conjugate base, acetate ion, thus:

CH₃CO₂H(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃CO₂⁻(aq )

Where dissociation constant, ka, is defined as the ratio of concentrations of products and reactants:

Ka = 1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

<em>H₂O is not taken into account in the equilibrium because is a pure liquid</em>

<em />

When a solution of acetic acid becomes to equilibrium, the original concentration of the acid decreases producing more H₃O⁺ and CH₃CO₂⁻.

The concentrations at equilibrium when a 0.100M solution of acetic acid reaches this state, is:

[CH₃CO₂H] = 0.100M - X

[H₃O⁺] = X

[CH₃CO₂⁻] = X

<em>Where X is reaction coordinate.</em>

Replacing in Ka expression:

1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

1.8x10⁻⁵ = [X] [X] / [0.100M - X]

1.8x10⁻⁶ - 1.8x10⁻⁵X = X²

1.8x10⁻⁶ - 1.8x10⁻⁵X - X² = 0

Solving for X:

X = -0.00135 → False solution. There is no negative concentrations.

X = 0.00133 → Right solution.

That means concentration of acetate ion is:

[CH₃CO₂⁻] = 0.00133M.

Now, percent ionization is defined as 100 times the ratio between weak acid that is ionizated, [CH₃CO₂⁻] = 0.00133M, per initial concentration of the acid, [CH₃CO₂H] = 0.100M. Replacing:

% Ionization = 0.00133M / 0.100M × 100 =

<em />

4 0

3 years ago

Why are chemists great for solving problems?

sertanlavr [38]

Because only those with logic (problem solving capabilities) go to chemistry.

4 0

3 years ago

The sodium isotope with 13 neutrons. Express your answer as an isotope

wariber [46]

Answer:

$\large\boxed{\large\boxed{^{24}_{11}Na}}$

Explanation:

The question is asking to write the <em>isotopic symbol </em>of the form $^A_ZX$ for the <em>sodium isotope with 13 neutrons</em>.

In general, the isotopic symbol in the given form shows:

The element's chemical symbol: X

A: mass number of the isotope, written as a superscript to the left of the element's simbol, and

A = number of protons + number of neutrons

Z: atomic number of the isotope, written as a subscript to the left of the elements's symbol,

Z = number of protons

The atomic symbol of sodium is Na.

The atomic number, or number of protons, is the same for every isotope of the element, and you can find it in any periodic table. Tha atomic number of sodium is 11. Thus:

Z = 11

The mass number is the number of protons plus neutrons, hence:

A = 11 + 13 = 24

Now you can write the isotope symbol for the sodium isotope with 13 neutrons:

$^{24}_{11}Na$

6 0

4 years ago

The specific heat capacity (c) of a substance is the thermal energy required

Ierofanga [76]

Temperature of the same cup of water will rise by 6 °C unless it boils.

<h3>Explanation</h3>

$Q = c\cdot m \cdot \Delta T$ .

However, neither $c$ nor $m$ is given.

Adding $Q = 125\;\text{kJ}$ to this cup of water of mass $m$ rises its temperature by $\Delta T = 1 \; \textdegree{}\text{C}$ .

In other words,

$\begin{array}{lll}Q &=& c \cdot m \cdot \Delta T\\125\;\text{kJ} &=& c \cdot m \times (1 \; \textdegree{}\text{C})\end{array}$

$c \cdot m = \dfrac{Q}{\Delta T} = \dfrac{125\; \text{kJ}}{1 \; \textdegree{}\text{C}}$ .

Both $c$ and $m$ are constant for the same cup of water unless the water boils. It's possible to reuse the value of $c \cdot m$ in the second calculation. Here's how:

$\Delta T = \dfrac{Q}{c\cdot m} = \dfrac{750 \; \text{kJ}}{\dfrac{125\;\text{kJ}}{1\;\textdegree{}\text{C}}} = \dfrac{750}{125} \; \textdegree{}\text{C}} = 6 \; \textdegree{}\text{C}}$ .

4 0

3 years ago