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mafiozo [28]
3 years ago
6

Suppose that the root-mean-square velocity vrms of water molecules (molecular mass is equal to 18.0 g/mol) in a flame is found t

o be 1150 m/s. What temperature does this represent
Chemistry
1 answer:
Zolol [24]3 years ago
4 0

Answer:

T=954.41\ K

Explanation:

The expression for the root mean square speed is:

C_{rms}=\sqrt {\dfrac {3RT}{M}}

R is Gas constant having value = 8.314 J / K mol  

M is the molar mass of gas

Molar mass of water vapor = 18.0 g/mol  = 0.018 kg/mol

Temperature = ?

C_{rms}=1150  m/s

1150=\sqrt{\frac{3\times 8.314\times T}{0.018}}

1322500=\frac{24.942T}{0.018}

T=954.41\ K

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A 1.00 liter solution contains 0.43 M hydrofluoric acid and 0.56 M potassium fluoride. If 0.280 moles of potassium hydroxide are
kolbaska11 [484]

Answer:

Answers are in the explanation

Explanation:

Equlibrium of HF in H₂O is:

HF + H₂O ⇄ F⁻ + H₃O⁺

Now, the KOH reacts with HF, thus:

KOH + HF →  F⁻ + H₂O

<em>That means after reaction, concentration of HF decrease increasing F⁻ concentration.</em>

Now, seeing the equilibrium, as moles of HF decrease and F⁻ moles increase, the equilibrium will shift to the left decreasing H₃O⁺ concentration.

For the statements:

A. The number of moles of HF will increase. <em>FALSE</em>. HF react with KOH, thus, moles of HF decrease

B. The number of moles of F- will decrease. <em>FALSE</em>. The reaction produce  F⁻ increasing its moles.

C. The equilibrium concentration of H₃O⁺ will increase. <em>FALSE. </em>The equilibrium shift to the left decreasing concentration of H₃O⁺

D. The pH will decrease. <em>FALSE</em>. As the H₃O⁺ concentration decrease, pH will increase

E. The ratio of [HF] / [F-] will remain the same. <em>FALSE</em>. Because moles of HF are decreasing whereas F- moles are increasing changing, thus, ratio.

6 0
3 years ago
Based on the diagram below, how much of the excess reactant is left over? *
Alexxandr [17]

Answer:

3 of lunchmeat and 2 slices of cheese

Explanation:

From the question given,

Each sandwich contains:

2 bread + 3 lunch meat + 1 cheese.

Now, the limiting reactant is the slice of bread.

We can determine the leftover as follow:

1. For the lunchmeat.

From the simple equation above,

2 bread requires 3 lunchmeat.

Therefore, 6 bread will require = (6 x 3)/2 = 9 lunchmeat.

Lunchmeat given = 12

Lunchmeat required = 9

Leftover lunchmeat = 12 – 9 = 3

Therefore, 3 lunchmeat is leftover.

2. For cheese.

From the simple equation above,

2 bread requires 1 cheese.

Therefore, 6 bread will require = 6/2 = 3 cheese.

Cheese given = 5

Cheese required = 3

Leftover cheese = 5 – 3 = 2

Therefore, 2 cheese is leftover.

From the simple illustrations above,

3 lunchmeat and 2 cheese are leftover.

7 0
3 years ago
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