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Jet001 [13]
2 years ago
8

Ralph and Sheila are at a construction site and they are stacking bricks. The diagram above shows the bricks before they touch.

Use the information in the diagram to answer the question.
After the bricks have been touching for a while, whose top brick will be cooler?
Chemistry
2 answers:
noname [10]2 years ago
7 0

Answer:Sheila's top brick will be cooler than Ralph’s top brick, because Sheila’s started with more total energy, so less energy had to transfer for both her bricks to reach the same total energy.

Explanation:

aliya0001 [1]2 years ago
6 0

Answer: Sheila's top brick will be cooler than Ralph’s top brick, because Sheila’s started with more total energy, so less energy had to transfer for both her bricks to reach the same total energy.

Explanation:

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A. A pure chemical substance consisting of one type of atom
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What is the volume of a sample of ammonia gas (NH3) if it contains 0.23 moles?​
LekaFEV [45]

The volume of a sample of ammonia gas : 5.152 L

<h3>Further explanation</h3>

Given

0.23 moles of ammonia

Required

The volume of a sample

Solution

Assumed on STP

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

So for 0.23 moles :

= 0.23 x 22.4 L

= 5.152 L

8 0
2 years ago
A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
Mashcka [7]

Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

(a)  Relation between K_{a} and pK_{a} is as follows.

                       pK_{a} = -log (K_{a})

Putting the values into the above formula as follows.

                      pK_{a} = -log (K_{a})

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, relation between pH and  pK_{a} is as follows.

              pH = pK_{a} + log\frac{[conjugate base]}{[acid]}

                     = 3.347+ log \frac{0.15}{0.12}

                    = 3.44

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = Molarity \times volume

                                            = 11.6 M \times 0.001 L

                                             = 0.0116 mol

In the given reaction, NO^{-}_{2} will react with H^{+} to form HNO_{2}

Hence, before the reaction:

No. of moles of NO^{-}_{2} = 0.15 M \times 1.0 L

                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

                               = (0.12 + 0.0116) mol

                                = 0.1316 mol

As, K_{a} = 4.5 \times 10^{-4}

           pK_{a} = -log (K_{a})

                         = -log(4.5 \times 10^{-4})

                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = pK_{a} + log \frac{[conjugate base]}{[acid]}

            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

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