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3 years ago

How is energy transfer connected to your life

Chemistry

1 answer:

lutik1710 [3]3 years ago

4 0

Answer:

Baking, microwave, heating system for your house, water boiler, fridge.

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What is the mass of 3.0 mole of N20?

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The mass is 132.0384 grams.

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3 years ago

Which option lists the layers of the rainforest in the correct order from top to bottom?

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Answer: B.

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Trust me already took the quiz on it its B.

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3 years ago

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A sample of helium gas has a pressure of 1.5 atm at 23 C. At what Celsius

Nina [5.8K]

Answer:

180.87 °C

Explanation:

Applying pressure law,

P/T = P'/T'..................... Equation 1

Where P = Initial pressure of helium gas, T = Initial Temperature of helium gas, P' = Final Pressure of helium gas, T' = Final Temperature of helium gas

Make T' the subject of equation 1

T' = P'T/P......................... Equation 2

From the question,

Given: P = 1.5 atm, P' = 2.3 atm, T = 23°C = (23+273)K = 296 K

Substitute these values into equation 2

T' = (2.3×296/1.5)

T' = 680.8/1.5

T' = 453.87 K

T' = 453.87-273

T' = 180.87°C

3 0

3 years ago

The Ksp of AgI is 8.3× 10–17. You titrate 25.00 mL of 0.08160 M NaI with 0.05190 M AgNO3. Calculate pAg after the following volu

11111nata11111 [884]

Answer:

(a) 13.64; (b) 8.04; (c) 2.25

Explanation:

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

$K_{\text{sp}} = {\text{[Ag$^{+}$][I$^{-}$]} = 8.3\times 10^{-17}$

(a) pAg at 35.10 mL

$\text{Moles of I$^{-}$} = \text{0.02500 L} \times \dfrac{\text{0.08160 mol}}{\text{1 L}} = 2.040 \times 10^{-3}\text{ mol/L }\\\text{Moles of Ag$^{+}$} = \text{0.03510 L} \times \dfrac{\text{0.05190 mol}}{\text{1 L}} = 1.822 \times 10^{-3}\text{ mol/L}$

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

I/mol: 1.822 × 10⁻³ 2.040 × 10⁻³

C/mol: -1.822 × 10⁻³ -1.822 × 10⁻³

E/mol: 0 0.218 × 10⁻³

We have a saturated solution of AgI containing 0.218 × 10⁻³ mol of excess I⁻.

V = 25.00 mL + 35.10 mL = 60.10 mL

$\text{[I$^{-}$]} = \dfrac{0.218 \times 10^{-3}\text{ mol}}{\text{0.0610 L}} = 3.57 \times 10^{-3}\text{ mol/L}\\$

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

E/mol·L⁻¹: s 3.57 × 10⁻³ + s

$K_{\text{sp}} = s(3.57 \times 10^{-3} + s) = 8.3\times 10^{-17}\\$

Check for negligibility:

$\dfrac{3.57 \times 10^{-3}}{8.3\times 10^{-17}} = 4.3 \times 10^{13} > 400\\\\\therefore s \ll 3.63 \times 10^{-3}\\K_{\text{sp}} = s\times 3.63 \times 10^{-3}= 8.3\times 10^{-17}\\\\s = \text{[Ag$^{+}$]} = \dfrac{8.3\times 10^{-17}}{3.63 \times 10^{-3}} =2.29 \times 10^{-14}\\\\\text{pAg} = -\log \left (2.29\times 10^{-14} \right) = \mathbf{13.64}$

(b) At equilibrium

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

E/mol·L⁻¹: s s

$K_{\text{sp}} = s\times s = s^{2} = 8.3\times 10^{-17}\\s = \sqrt{8.3\times 10^{-17}} = 9.11 \times 10^{-9}\\\text{pAg} = -\log \left (9.11 \times 10^{-9} \right) = \mathbf{8.04}$

(c) At 47.10 mL

$\text{Moles of Ag$^{+}$} = \text{0.04710 L} \times \dfrac{\text{0.05190 mol}}{\text{1 L}} = 2.444 \times 10^{-3}\text{ mol}$

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

I/mol: 2.444 × 10⁻³ 2.040 × 10⁻³

C/mol: -2.040 × 10⁻³ -2.040 × 10⁻³

E/mol: 0.404 × 10⁻³ 0

V = 25.00 mL + 47.10 mL = 72.10 mL

$\text{[Ag$^{+}$]} = \dfrac{0.404 \times 10^{-3}\text{ mol}}{\text{0.0721 L}} = 5.61 \times 10^{-3}\text{ mol/L}\\\text{pAg} = -\log(5.61 \times 10^{-3}) = \mathbf{2.25}$

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4 years ago

Help!!will give brainliest

Aleksandr-060686 [28]

Answer:

As a pyramid gets taller, it gets smaller. In the Linnaean classification system, this happens as well, with each category growing smaller.

Explanation:

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3 years ago