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2 years ago

For a particular experiment, Nadia must measure the following quantities:

Chemistry

2 answers:

rjkz [21]2 years ago

8 0

Answer:

Graduated cylinder

Stopwatch

Thermometer

Electronic Balance

Explanation:

The answers on Edge are above. (VERIFIED Edge Answer)

Make sure to click "thanks" and rate this answer to let others know this is the correct answer for this question.

Edge Tip: Answers do not mix or change.

g100num [7]2 years ago

4 0

Graduated cylinder
Stopwatch
Thermometer

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In the following reaction, how many grams of NaBr will react with 311 grams of Pb(NO3)2?

Mashutka [201]

Molar mass :

NaBr = 103 g/mol

Pb(NO3)2 = 331.20 g/mol

Balanced chemical equation :

2 NaBr + 1 Pb(NO3)2 = 2 NaNO3 + 1 PbBr2

2*103 g NaBr ------------> 1 * 331.20 g Pb(NO3)2
g NaBr -------------------> 311 g Pb(NO3)2

331.20 g = 2*103*311

331.20 g = 64066

mass ( NaBr ) = 64066 / 331.20

mass ( naBr) = 193,43 g of NaBr

hope this helps!.

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2 years ago

Drag the tiles to the correct boxes to complete the pairs.

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Answer:

Explanation:

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2 years ago

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2 years ago

What layer of earths atmosphere is the layer next to the earth

Pavlova-9 [17]

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2 years ago

A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )

Aleksandr [31]

Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)

The reaction rate of equation (1) is given by:

$rate = k*[O_{3}][NO]$ (2)

We have:

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

$rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s$

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

$\frac{\Delta[NO_{2}]}{\Delta t} = rate$

$\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}$

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!

5 0

2 years ago