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Nadya [2.5K]
2 years ago
7

the density of gold is 19.3 g/cm^3. You inherit a gold bar that measures 7.0 in by 3.0 in by 2.0 in. How many ounces does the ba

r weigh? Express answer in scientific notation.
Chemistry
1 answer:
podryga [215]2 years ago
8 0
<span>You can work out the following: Mass = Density x Volume Density = Mass Ă· Volume Volume = Mass Ă· Density The density of gold is 0.670205 oz/cm^3 The volume is length x width x depth = 7*3*2 = 42 The answer is Mass = 0.670205 x 42 = 281 ounces</span>
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What will happen to the pH of a solution when the [H+] is increased?
matrenka [14]

Answer:

decreases

Explanation:

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8 0
3 years ago
If 0.750 L of argon at 1.50 atm and 177°C and 0.235 L of sulfur dioxide at 95.0 kPa and 63.0°C are added to a 1.00-L flask and t
Furkat [3]

Answer:

The resulting pressure in the flask is 0.93 atm

Explanation:

- Apply the Ideal Gas law in both cases to get the mols, of Ar and SO2.

- Once you know the mols, sum both to get the total mols in the mixture.

- Apply the Ideal Gas lawin the flask with the total mols to know the resulting pressure.

First: 0.750 L of argon at 1.50 atm and 177°C

T° C + 273 = T° K → 177°C + 273 = 450K

P .V = n . R . T

1.50 atm . 0.750 L = n . 0.082 L.atm/mol.K  . 450K

(1.50 atm . 0.750 L) /  (0.082 mol.K/L.atm  . 450K) = n

0.030 mols Ar = n

Be careful with the R units, the ideal gases constant

Let's convert kPa to atm.

101.33 kPa _____ 1 atm

95 kPa ________ (95 / 101.33) = 0.94 atm

T° C + 273 = T° K → 63°C + 273 = 336 K

0.94 atm . 0.235 L = n . 0.082 L.atm/mol.K . 336K

(0.94 atm . 0.235 L) / (0.082 mol.K/L.atm . 336K) = n

8.01X10⁻³ mols = n

0.030 mols Ar  + 8.01X10⁻³ mols SO₂ = 0.038 total mols in the mixture

1L . P = 0.038 mol . 0.082 L.atm / mol.K . 298 K

P = (0.038 mol . 0.082 L.atm / mol.K . 298 K ) / 1L

P = 0.93 atm

7 0
2 years ago
He equation for the dissociation of pyridine is
sergeinik [125]

Answer:

10.10

Explanation:

Step 1: Write the basic dissociation reaction for pyridine

C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq)      Kb = 1.9 × 10⁻⁹

Step 2: Calculate [OH⁻]

For a weak base, we will use the following expression.

[OH⁻] = √(Cb × Kb) = √(9.2 × 1.9 × 10⁻⁹) = 1.3 × 10⁻⁴ M

Step 3: Calculate pOH

We will use the definition of pOH.

pOH = -log [OH⁻] = -log 1.3 × 10⁻⁴ = 3.9

Step 4: Calculate pH

We will use the following expression.

pH = 14 - pOH = 14 - 3.9 = 10.10

3 0
3 years ago
Determine the percentage of carbon and hydrogen in ethane C2H6 if the molecular weight is 30.
Naddika [18.5K]

Answer:

Percentage of carbon:

{ \tt{ =  \frac{24}{30}  \times 100\%}} \\  = 80\%

Percentage of hydrogen:

{ \tt{ =  \frac{6}{30}  \times 100\%} } \\  = 20\%

8 0
3 years ago
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