A company charges $5.00 to rent a paddle boat for the first hour, plus $2.00 for each additional hour.

How are compounds with ionic and covalent bonds alike?

PSYCHO15rus [73]

Answer:

B

Explanation:

8 0

4 years ago

Which band is the smallest within the electromagnetic spectrum? A) X rays B) microwaves C) radio waves Eliminate D) visible ligh

aev [14]

Radio wave eliminate band (answer c) is smallest within electromagnetic spectrum.

The electromagnetic spectrum is the range of frequencies (the spectrum) of electromagnetic radiation and their respective wavelength and photon energies. The electromagnetic spectrum range from radio waves to gamma rays.Radio waves has the shortest wavelength while gamma rays has highest.

4 0

3 years ago

Read 2 more answers

Two trials are run, using excess water. In the first trial, 7.8 g of Na2O2(s) (molar mass 78 g/mol) is mixed with 3.2 g of S(s).

Tema [17]

Answer:

The answer is "Option C".

Explanation:

Given equation:

$2Na_20_2 (s)+S(s)+2H_2O \longrightarrow 4NaOH(aq)+SO_2(aq)$

$\to \Delta H^{\circ}_{rxn} (298\ K) = -610 \frac{kJ}{mol}$

$\to Na_2O_2 \ Mass = 7.8 \ g\\\\ \to Na_2O_2 \ Molar \ mass = 78 \frac{g}{mol}$

$Na_2O_2$ Has been the reactant which is limited since the two experiments are equal to $Na_2O_2$ for relationship between stress amounts.

$Na_2O_2, n =\frac{Mass of Na_2O_2}{Molar mass of Na_2 O_2}=\frac{7.8 \ g}{78 \frac{g}{mol}} =0.1 \ mol \\\\q=\Delta H^{\circ}_{rxn} \times n = \frac{ -610 \ kJ}{ 2 \ mol \ Na_2 O_2} \times 0.1 \ mol \ Na_2O_2= 30.5 \ KJ\\\\$

Limiting reactant = $Na_2O_2$

$q=30.5 \ kJ \approx 30 \ kJ$

8 0

3 years ago

A 2.350×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then b

sveta [45]

Answer:

Part A: m = 0.02356 mol/kg = 0.02356 m

Part B: Xsolute = 4.243×10⁻⁴

Part C: % m/m = 0.1376%

Part D: ppm = 1,376 ppm

Explanation:

1. Data:

a) M = 2.350×10⁻² M

b) V sol = 1.000 L

c) V H₂O = 994.4 mL = 0.9944 L

d) d H₂O = 0.9982 g/mL

2. Formulae

M = n solute / V sol (L)
m = n solute / Kg solvent
X solute = n solute / N total
% m/m = (mass of solute / mass of solution) × 100
ppm = (mass of solute / mass of solution) × 1,000,000
density = mass in grams / volume in mL

3. Solution

Part A: Calculate the molality of the salt solution.

m = n solute / Kg solvent

i) M = n solute / V sol (L) ⇒ n solute = M × V sol (L)

⇒ n solute = M = 2.350×10⁻² M × 1.000 L = M = 2.350×10⁻² mol

ii) density H₂O = mass H₂O / volume H₂O

⇒ mass H₂O = density H₂O × volume H₂O

⇒ mass H₂O = 0.9982 g/mL × 999.4 mL = 997.6 g

iii) kg H₂O = 997.6 g / (1,000 g/Kg) = 0.9976 kg

iv) m = 2.350×10⁻² mol / 0.9976 kg = 0.02356 mol/kg = 0.02356 m

Part B: Calculate the mole fraction of salt in this solution.

X solute = n solute / N total

i) n solute = 2.350×10⁻² mol

ii) n solvent = n H₂O = mass H₂O in grams/ molar mass H₂O

⇒ 997.6 g / 18.015 g/mol = 55.38 mol

iii) X solute = 2.350×10⁻² mol / 55.38 mol = 4.243×10⁻⁴

Part C: Calculate the concentration of the salt solution in percent by mass.

% m/m = (mass of solute / mass of solution) × 100

i) molar mass = mass in grams / molar mass

⇒ mass of solute = mass of NaCl = n solute × molar mass NaCl

⇒ mass of solute = 2.350×10⁻² mol × 58.44 g/mol = 1.373 g

ii) % m/m = (1.373 g / 997.6 g) × 100 = 0.1376%

Part D: Calculate the concentration of the salt solution in parts per million.

ppm = (mass of solute / mass of solution) × 1,000,000

i) ppm = ( (1.373 g / 997.6 g) × 1,000,000 = 1,376 ppm

5 0

3 years ago

An 18 liter container holds 16.00 grams of oxygen gas (o2 at 45 °c. what is the pressure in the container?

Scrat [10]

Hello!

The pressure of an 18 L container which holds 16,00 grams of oxygen gas (O₂) at 45 °C is 0,725 atm

To solve this problem we first need to set up the data in the appropriate units to input it in the Ideal Gas Law.

a) 16 g of Oxygen gas to moles of oxygen gas:

$16gO_2* \frac{1 mol O_2}{32gO_2}=0,5 mol O_2$

b) 45 °C to K

$K=$^{\circ}$C + 273,15 = 45 $^{\circ}$C + 273.15=318,15 K$

Now, we clear the Ideal Gas Equation for P, and solve it:

$P*V=n*R*T \\ \\ P= \frac{n*R*T}{V}= \frac{(0,5mol)*(0,082 \frac{L*atm}{mol*K})* (318,15 K)}{18 L}= 0,725 atm$

Have a nice day!

4 0

3 years ago