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3 years ago

Which element in the list is the most reactive?

Chemistry

2 answers:

Alina [70]3 years ago

3 0

Answer:

Cs (Cesium)

Explanation:

...

Reil [10]3 years ago

3 0

Answer:

Cs (Cesium)

Element number 35 and group 1

Explanation:

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What is the pH at the equivalence point in the titration of a 25.7 mL sample of a 0.370 M aqueous nitrous acid solution with a 0

expeople1 [14]

Answer:

pH = 8.24

Explanation:

Nitrous acid (HNO₂) reacts with KOH, thus:

HNO₂ + KOH → KNO₂ + H₂O

Moles of HNO₂ are:

0.0257mL ₓ (0.370mol / L) = 0.00951moles.

In equivalence point, the complete moles of nitrous acid reacts with KOH producing potassium nitrite. There are needed:

0.00951mol ₓ (1L / 0.491mol) = 0.01937L ≡ 19.4mL of 0.491M KOH to reach equivalence point.

Total volume in equivalence point is: 19.4mL + 25.7mL = 45.1mL

Potassium nitrite is in equilibrium with water, thus:

NO₂⁻ + H₂O ⇄ HNO₂ + OH⁻

Where equilibrium constant, Kb, is defined as:

Kb = 1.41x10⁻¹¹ = $\frac{[OH^-][HNO_2]}{[NO_2]}$

In equilibrium, molarity of each compound are:

[NO₂⁻]: 0.00951mol/0.00451L - X = 0.211M - X

[HNO₂]: X

[OH⁻]: X

Where X is reaction coordinate

Replacing in Kb:

1.41x10⁻¹¹ = $\frac{[X][X]}{[0.211 -X]}$

0 = X² + 1.41x10⁻¹¹X - 2.97x10⁻¹²

Solving for X:

X = -1.72x10⁻⁶ FALSE ANSWER. There is no negative concentrations.

X = 1.72x10⁻⁶. Right answer.

That means:

[OH⁻]: 1.72x10⁻⁶M

As pOH is -log [OH⁻] and pH = 14-pOH:

pOH = 5.76; pH = 8.24

3 0

4 years ago

Given the balanced equation:

Aliun [14]

Answer:

0.4

Explanation:

Given data:

Number of moles of SrCl₂ consumed = ?

Mass of ZnCl₂ produced = 54 g

Solution:

Chemical equation:

ZnSO₄ + SrCl₂ → SrSO₄ + ZnCl₂

Number of moles of ZnCl₂:

Number of moles = mass/ molar mass

Number of moles = 54 g/136.3 g/mol

Number of moles = 0.4 mol

Now we will compare the moles of ZnCl₂ with SrCl₂ from balance chemical equation.

ZnCl₂ : SrCl₂

1 : 1

0.4 : 0.4

Thus when 54 g of ZnCl₂ produced 0.4 moles of SrCl₂ react.

6 0

4 years ago

Cylinder of air at 1.5 atm of pressure is kept at room temperature while a piston compresses the air from 40 l down to 10 ml. wh

djyliett [7]

The new pressure, P₂ is 6000 atm.

<h3>Calculation:</h3>

Given,

P₁ = 1.5 atm

V₁ = 40 L = 40,000 mL

V₂ = 10 mL

To calculate,

P₂ =?

Boyle's law is applied here.

According to Boyle's law, at constant temperature, a gas's volume changes inversely with applied pressure.

PV = constant

Therefore,

P₁V₁ = P₂V₂

Put the above values in the equation,

1.5 × 40,000 = P₂ × 10

P₂ = 1.5 × 4000

P₂ = 6000 atm

Therefore, the new pressure, P₂ is 6000 atm.

Learn more about Boyle's law here:

brainly.com/question/23715689

#SPJ4

8 0

2 years ago

B. 10% is the correct answer

GarryVolchara [31]

No the right answer is c

5 0

3 years ago

Read 2 more answers

For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a

Elza [17]

Answer:

The volume decreases 5.5%

Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

If this is an ideal gas, the equation to use is:

PV = nRT

Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

n₁ = n₂ = n

The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:

P₁V₁ = nRT₁ -----> n = P₁V₁ / RT₁

Doing the same with the pressure and volume 2 we have:

n = P₂V₂ / RT₂

Equalling both expressions and solving for V₂:

P₁V₁ / RT₁ = P₂V₂ / RT₂

V₂ = P₁T₂V₁ / P₂T₁

Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:

V₂ = P₁T₂V₁ / 0.95P₁T₁

The values of temperature in K:

T1 = 21+273 = 294 K

T2 = -8 + 273 = 265 K

Finally, let's calculate the volume:

V₂ = 264*P₁*V₁ / 294*0.95*P₁ ----> P cancels out

V₂ = 264V₁ / 294*0.95

V₁ = 0.945V₂

With this, we can day that Volume 2 decreases.

Now the percentage change would be using the following expression:

%V = (V₁ - V₂ / V₁) * 100

Replacing the data we have:

%V = V1 - 0.945V₁ / V₁

%V = 0.055V₁ / V₁ * 100

%V = 5.5%

7 0

3 years ago