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3 years ago

Is the equation above balanced or unbalanced? Explain? Worth 55 points.

Chemistry

2 answers:

Paul [167]3 years ago

8 0

Answer:

For me, it is balanced

Explanation:

Because the amount of element Zn reaction is equal the production. Also as the amount of Hydrogen and Chloride in the reaction and the production are equal.

Alona [7]3 years ago

7 0

Balanced even though it is formed differently on each side they are the same. When counting the coefficient in front of it they are still the same.

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Observe the activity in your classroom. You can observe people, objects in the room, parts of your actual classroom, to begin no

ICE Princess25 [194]

Example 1: Crushing a can.

Type of change:

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4 0

3 years ago

If a buffer contains 1.05M B and 0.750M BH+ has the pH of 9.5. What would be the pH after 0.005mol of HCL is added to 0.5L of so

Leviafan [203]

Answer:

Final pH: 9.49.

Round to two decimal places as in the question: 9.5.

Explanation:

The conjugate of B is a cation that contains one more proton than B. The conjugate of B is an acid. As a result, B is a weak base.

What's the pKb of base B?

Consider the Henderson-Hasselbalch equation for buffers of a weak base and its conjugate acid ion.

$\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}}$ .

$\text{pOH} = \text{pK}_w - \text{pH}$ .

$\text{pK}_w = 14$ .

$\text{pOH} = 14 - 9.5 = 4.5$

$\displaystyle \text{pK}_b = \text{pOH} -\log{\frac{[\text{Salt}]}{[\text{Base}]}}\\\phantom{\text{pK}_b} = 4.5 - \log{\frac{0.750}{1.05}} \\\phantom{\text{pK}_b} =4.64613$ .

What's the new salt-to-base ratio?

The 0.005 mol of HCl will convert 0.005 mol of base B to its conjugate acid ion BH⁺.

Initial:

$n(\text{B}) = c\cdot V = 1.05 \times 0.5 = 0.525\;\text{mol}$ ;
$n(\text{BH}^{+}) = c\cdot V = 0.750 \times 0.5 = 0.375\;\text{mol}$ .

After adding the HCl:

$n(\text{B}) = 0.525 - 0.005 = 0.520\;\text{mol}$ ;
$n(\text{BH}^{+}) = 0.375+ 0.005 = 0.380\;\text{mol}$ .

Assume that the volume is still 0.5 L:

$\displaystyle [\text{B}] = \frac{n}{V} = \frac{0.520}{0.5} = 1.04\;\text{mol}\cdot\text{dm}^{-3}$ .
$\displaystyle [\text{BH}^{+}] = \frac{n}{V} = \frac{0.380}{0.5} = 0.760\;\text{mol}\cdot\text{dm}^{-3}$ .

What's will be the pH of the solution?

Apply the Henderson-Hasselbalch equation again:

$\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}} = 4.64613 + \log{\frac{0.760}{1.04}} = 4.50991$

$\text{pH} = \text{pK}_w - \text{pOH}= 14 - 4.50991 = 9.49$ .

The final pH is slightly smaller than the initial pH. That's expected due to the hydrochloric acid. However, the change is small due to the nature of buffer solutions: adding a small amount of acid or base won't significantly impact the pH of the solution.

3 0

3 years ago

In the reaction of nitrogen with hydrogen to produce ammonia, NH3, how many moles of ammonia would be produced from 1.0 mol of h

Lina20 [59]

N2 + 3H2 ------> 2NH3
According to the equation;
3 mol of hydrogen will produce 2mol of ammonia.
1 mol of hydrogen _____________________?

= 2 × 1/3

= 0.67 mol

7 0

3 years ago

Read 2 more answers

A student is trying to calculate the density of a ball. She already knows the mass, but she needs to determine the volume as wel

Reptile [31]

Let's examine the given choices:
V = (4/3)*pi*(r)^3 is the rule to calculate the volume of the sphere
V = (1/3)*pi*(r)^2*h is the rule to calculate the volume of the cone
V = s^3 is the rule to calculate volume of the cube
V = pi*(r)^2*h is the rule to calculate volume of the cylinder

Now, the ball can be treated as a perfect sphere, therefore, to calculate its volume we will use the rule of the volume of sphere.

Based on this, the correct choice is:
V = (4/3)*pi*(r)^3

7 0

3 years ago

Read 2 more answers

Two liters of oxygen (O2) is collected over water at 20oC. If the total pressure in the container is 95.00 kPa, what is the part

garri49 [273]

Answer:

THE PARTIAL PRESSURE OF OXYGEN GAS IN THE CONTAINER IS 92.67kPa WHICH IS OPTION B.

Explanation:

To calculate the partial pressure of oxygen gas collected over water, we use

Ptotal = Poxygen + P water

It is worthy to note that when oxygen is collected over water, it is mixed with water vapor and the total pressure in the container will be the sum of the pressure exerted by the oxygen gas and that of the water vapor at that given temperature.

At 20 C, the vapor pressure of water as given in the question is 2.33 kPa.

Using the above formula,

Ptotal = Poxygen + P water

Substituting for Poxygen, we have;

Poxygen = Ptotal - P water vapor

P oxygen = 95 .00 kPa - 2.33 kPa

P oxygen = 92.67 kPa.

The partial pressure of oxygen gas in the container is hence, 92.67kPa.

3 0

3 years ago