All categories

3 years ago

Question 1 (1 point)

Chemistry

1 answer:

Colt1911 [192]3 years ago

8 0

Answer:

Both are made up of two substances that are not chemically combined.
oxygen (O2)
carbon dioxide (CO2 )
air (N2 mixed with O2 and CO2)
Pure substances cannot be separated by physical means.

Explanation:

You might be interested in

Where and when are acute-phase proteins produced?

mrs_skeptik [129]

Answer:

The answer should be C. Primarily in the liver in response to inflammation :)

Have an amazing day!!

Please rate and mark brainliest!!

3 0

2 years ago

In a solution of a carbonated beverage, what is the dissolved carbon dioxide?

insens350 [35]

Answer:

Explanation:

D is the correct answer because, in aqueous solution, solvent is water and solute (in this example carbon dioxide CO₂) is a substance dissolved in water. The amount of solute that can be dissolved in a solvent depends of chemical composition, temperature and pressure

4 0

3 years ago

Read 2 more answers

What is the mass of 61.9 L of oxygen gas collected at STP?

Tcecarenko [31]

Answer:

D. 44.2 g O₂

General Formulas and Concepts:
<u>Gas Laws</u>

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at <em>1 atm, 273 K</em>

<u>Stoichiometry</u>

Dimensional Analysis
Mole Ratio

Explanation:

<u>Step 1: Define</u>

<em>Identify given.</em>

61.9 L O₂ at STP

<u>Step 2: Convert</u>

We know that the oxygen gas is at STP. Therefore, we can set up and solve for how many <em>moles</em> of O₂ is present:

$\displaystyle 61.9 \ \text{L} \ \text{O}_2 \bigg( \frac{1 \ \text{mol} \ \text{O}_2}{22.4 \ \text{L} \ \text{O}_2} \bigg) = 2.76339 \ \text{mol} \ \text{O}_2$

Recall the Periodic Table (Refer to attachments). Oxygen's atomic mass is roughly 16.00 grams per mole (g/mol). We can use a mole ratio to convert from <em>moles</em> to <em>grams</em>:

$\displaystyle 2.76339 \ \text{mol} \ \text{O}_2 \bigg( \frac{16.00 \ \text{g} \ \text{O}_2}{1 \ \text{mol} \ \text{O}_2} \bigg) = 44.2143 \ \text{g} \ \text{O}_2$

Now we deal with sig figs. From the original problem, we are given 3 significant figures. Round your answer to the <u>exact</u> same number of sig figs:

$\displaystyle 44.2143 \ \text{g} \ \text{O} \approx \boxed{ 44.2 \ \text{g} \ \text{O}_2 }$

∴ our answer is letter choice D.

---

Topic: AP Chemistry

Unit: Stoichiometry

6 0

2 years ago

A student mixes in a test tube 3.00mL of 0.050M CuSO4with 7.00mL of 0.20M NH3/NH41 . The solution becomes a deep blue color. Ass

valkas [14]

Answer:

$\large \boxed{\text{0.0035 mol/L}}$

Explanation:

We are given the volumes and concentrations of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

Cu²⁺ + 4NH₃ ⟶ Cu(NH₃)₄²⁺

V/mL: 3.00 7.00

c/mol·L⁻¹: 0.050 0.20

1. Identify the limiting reactant

(a) Calculate the moles of each reactant

$\text{Moles of Cu}^{2+}= \text{3.00 mL solution} \times \dfrac{\text{0.050 mmol Cu}^{2+}}{\text{1 mL solution}} = \text{0.150 mmol Cu}^{2+}\\\\\text{Moles of NH}_{3} = \text{7.00 mL solution} \times \dfrac{\text{0.20 mmol NH}_{3}}{\text{1 mL solution}} = \text{0.140 mmol NH}_{3}$

(b) Calculate the moles of Cu(NH₃)₄²⁺ that can be formed from each reactant

(i) From Cu²⁺

$\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.150 mmol Cu}^{2+} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{1 mmol Cu}^{2+}}\\\\= \text{0.150 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}$

(ii) From NH₃

$\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.140 mmol NH}_{3} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{4 mmol NH}_{3}}\\\\= \text{0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}$

NH₃ is the limiting reactant, because it forms fewer moles of the complex ion.

(c) Concentration of the complex ion

$\text{The reaction forms 0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$ in a total volume of 10.00 mL.}\\c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.0350 mmol}}{\text{10.00 mL}} = \textbf{0.0035 mol/L}\\\\\text{The concentration of the complex ion is $\large \boxed{\textbf{0.0035 mol/L}}$}$

7 0

3 years ago

How many particles are in 1.40 x 10^7 mol of aluminum (Al)?

SpyIntel [72]

Answer:

<h3>Hlo there !! </h3>

<u>One mole of any substance contains 6.022*1023 structural units (atoms, molecules, ions, etc.). This number is known as the Avogadro constant.</u>

<u>One mole of any substance contains 6.022*1023 structural units (atoms, molecules, ions, etc.). This number is known as the Avogadro constant.So 1.04*107 mol of Al contains 1.40*107 * 6.022*1023 = 8.43*1030 structural units (in case of Al – atoms).</u>

<h3><u>8.43*1030 particles Al.</u></h3>

Explanation:

<h3>Hope this helps !!</h3>

5 0

2 years ago