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natulia [17]
3 years ago
13

Why are we able to measure the atmospheric pressure in the room and equate this to the pressure of a gas in a test tube?

Chemistry
1 answer:
laiz [17]3 years ago
5 0
Atmospheric pressure is measured with a barometer. A barometer consists of an evacuated vertical tube with its top end closed and its bottom end resting in a container of mercury that is open to the atmosphere The pressure exerted by the atmosphere acts on the exposed surface of the liquid to force mercury up into the tube. Sea level atmospheric pressure will support a mercury column generally not more than 29.92-in. high. Thus, the standard for atmospheric pressure at sea level is 29.92 in.-Hg, which translates to an absolute pressure of 14.69 psia.
The two basic reference points in all these measurements are standard atmospheric pressure and a perfect vacuum.
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7. A solution containing 90grams of KNO3 per 100. grams of H2O at
Tomtit [17]

Answer:

(4) concentrated and supersaturated

Explanation:

At 50.°C, 90g of KNO3 lies above the solubility curve [on the Regents Reference Table G]. This indicates that the solution is supersaturated, meaning it contains more solute than will naturally dissolve, and was formed when a saturated solution cooled. Furthermore, the percent concentration of this solution is 90% KNO3 making this solution concentrated. This can be calculated using the formula for mass percent concentration.

Percent Mass = <u>Mass of Solute (g)</u> x 100

                         Mass of Solution (g)

5 0
3 years ago
1. How many grams of potassium chloride (KCI)<br> can be dissolved at 10°C?
Allisa [31]

Answer:

At 10 ∘C , the solubility of potassium chloride is measured to be 31.2⋅g per 100⋅g of water.

Explanation:

3 0
3 years ago
PLEASE HELP ME THIS IS DUE RN I HAVE NO IDEA I WILL MARK BRAINLIEST I JUST NEED HELP
djverab [1.8K]

Answer:

3.45 × 10^5

=3.45^5

I hope this helps

6 0
3 years ago
When we have two sorted lists of numbers in non-descending order, and we need to merge them into one sorted list, we can simply
Dovator [93]

Answer:

Explanation:

Firstly of all we have to construct the min-heap of the k-sub list and each sub list which is a node in the constructed min-heap.

We have several steps to follows:

Step-1. When we compare the two sub lists, at the starting we can compare their first elements which is actually their minimum elements.

Step-2. The min-heap formation will cost be O(k) time.

Step-3. After the step 1 & step 2 we can run the minimum algorithm which can be extracted from the minimum element in the root list.

Step-4. Then Update the root list in the heap and after that simplify the min-heap as maintained by the new minimum element in the root list.

Step-5. If any root sub-list becomes empty in the step 4 then we can take any leaf sub-list from the root and simplify it.

Step-6. At every Extraction of the element it can take up to O(log k) time.

Hence, We can say that the extract of n element in the total whose

Running time will be O(n log k + k) which can be equal to the O(n log k+ k) (since k < n).

7 0
3 years ago
Complete each of the following questions by choosing the correct words
suter [353]

Answer:

to the first

A) mass

to the second

B) Volume

7 0
2 years ago
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