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Stella [2.4K]
3 years ago
13

Calculate the root mean square velocity of gaseous argon atoms at 26 ∘c.

Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
5 0
By substitution in this formula :
V= √3RTM
when R = 8.31451 J.mol^-1K^-1
T is the tempreture in K = 26 °C + 273 = 299 k
M is the molar mass of argon Kg= 39.9 g = 39.9/1000= 0.0399 Kg
∴V = √3*8.31451 * 299 / 0.0399 
∴ V = 432 m/s 
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5 0
3 years ago
A mixture contains only nacl and al2(so4)3. a 1.45 g sample of the mixture is dissolved in water, and an excess of naoh is added
Aleks [24]

When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced.  The balanced chemical reaction is represented as-

Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄

On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.

As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.

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7 0
3 years ago
How much energy (in Joules) would it take to warm 3.11 grams of gold by 7.7 oC
Reptile [31]

Answer:

It would take 3.11 J to warm 3.11 grams of gold

Explanation:

Step 1: Data given

Mass of gold = 3.11 grams

Temperature rise = 7.7 °C

Specific heat capacity of gold = 0.130 J/g°C

Step 2: Calculate the amount of energy

Q = m*c*ΔT

⇒ Q = the energy required (in Joules) = TO BE DETERMINED

⇒ m = the mass of gold = 3.11 grams

⇒ c = the specific heat of gold = 0.130 J/g°C

⇒ ΔT = The temperature rise = 7.7 °C

Q = 3.11 g * 0.130 J/g°C * 7.7 °C

Q = 3.11 J

It would take 3.11 J to warm 3.11 grams of gold

4 0
3 years ago
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Sav [38]

Answer:

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