Which set of quantum numbers cannot specify an orbital?.

Jill is making lemonade. She mixes together some water, ice, lemon juice, and sugar. The lemonade is classified as a(n) ________

alina1380 [7]

The answer is mixture I think

5 0

4 years ago

Read 2 more answers

A solution of a concentration of H+ (10-4 M) has a pH of

Lesechka [4]

Answer:

pH = 4

Explanation:

Step 1: Given data

Concentration of H⁺ ions in the solution ([H⁺]): 10⁻⁴ M

Step 2: Calculate the pH of the solution

We will use the definition of pH.

pH = -log [H⁺]

pH = -log 10⁻⁴ M

pH = 4

The pH of the solution is 4. Considering the pH scale, given the pH is lower than 7, the solution is acidic.

8 0

3 years ago

Calculate the molality of a solution containing 257 g glucose (c6h12o6) dissolved in 1.62 l of water. assume a density of 1.00 g

Alex777 [14]

Molality is defined as the number of moles of solute in 1 kg of solvent.
since density and volume has been given we can calculate the mass of water
mass = density x volume
mass = 1.00 g/mL x 1620 mL = 1620 g
number of moles of glucose - 257 g / 180 g/mol = 1.43 mol
number of moles of glucose in 1620 g - 1.43 mol
therefore number of moles in 1000 g - 1.43 mol / 1.620 kg = 0.883 mol/kg
therefore molality is 0.883 mol/kg

4 0

4 years ago

Question 1(Multiple Choice Worth 4 points)

zubka84 [21]

<u>Answer </u>

Answer 1 : 28.9 g of CO is needed.

Answer 2 : Six moles of $H_{2}O$ over Nine moles of $O_{2}$

Answer 3 : Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$ .

Answer 4 : Mass of $O_{2}$ = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Answer 5 : 8.4 moles of sodium cyanide (NaCN) would be needed.

<u>Solution </u>

Solution 1 : Given,

Given mass of $Fe_{2}O_{3}$ = 55 g

Molar mass of $Fe_{2}O_{3}$ = 159.69 g/mole

Molar mass of CO = 28.01 g/mole

Moles of $Fe_{2}O_{3}$ = $\frac{\text{ Given mass of } Fe_{2}O_{3}}{\text{ Molar mass of } Fe_{2}O_{3}}$ = $\frac{55 g}{159.69 g/mole}$ = 0.344 moles

Balanced chemical reaction is,

$Fe_{2}O_{3}(s)+3CO(g)\rightarrow 2Fe(s)+3CO_{2}(g)$

From the given reaction, we conclude that

1 mole of $Fe_{2}O_{3}$ gives → 3 moles of CO

0.344 moles of $Fe_{2}O_{3}$ gives → 3 × 0.344 moles of CO

= 1.032 moles

Mass of CO = Number of moles of CO × Molar mass of CO

= 1.032 × 28.01

= 28.90 g

Solution 2 : The balanced chemical reaction is,

$2C_{3}H_{6}+9O_{2}\rightarrow 6CO_{2}+6H_{2}O$

From the given reaction, we conclude that the Six moles of $H_{2}O$ over Nine moles of $O_{2}$ is the correct option.

Solution 3 : The balanced chemical reaction is,

$4Fe+3O_{2}\rightarrow 2Fe_{2}O_{3}$

From the given balanced reaction, we conclude that Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$ .

Solution 4 : Given,

Given mass of $Zn(ClO_{3})_{2}$ = 150 g

Molar mass of $Zn(ClO_{3})_{2}$ = 232.29 g/mole

Molar mass of $O_{2}$ = 31.998 g/mole

Moles of $Zn(ClO_{3})_{2}$ = $\frac{\text{ Given mass of }Zn(ClO_{3})_{2} }{\text{ Molar mass of } Zn(ClO_{3})_{2}}$ = $(\frac{150\times 1}{232.29})moles$