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ipn
1 month ago
10

Elements with atomic numbers of 104 and greater are known as super-heavy elements. None of these elements have been found in nat

ure but instead have been made in a laboratory. They are very difficult and expensive to create, and they break down into other elements quickly. There currently are no practical applications for any of these elements. However, some scientists believe that, with further work, they may discover some isotopes of super-heavy elements that are more stable and that could possibly have practical implications. Do you think that scientists should continue to try to create super-heavy elements and expand the periodic table? Explain why or why not.
Chemistry
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A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o
Anarel [2989]

Response:

702 mL

To elaborate:

Given the following:

Mass of sodium hydroxide = 13.20 g

Molarity of H₂SO₄ = 0.235 M

We're tasked with finding the volume of acid necessary to neutralize the sodium hydroxide solution

Step 1: Write the balanced reaction equation

The reaction between H₂SO₄ and NaOH can be summarized as follows:

2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

Step 2: Calculate the moles of NaOH

Moles are calculated by dividing mass by molar mass

The molar mass of NaOH is 40.0 g/mol

Hence;

Number of moles of NaOH = 13.20 g ÷ 40 g/mol

= 0.33 moles of NaOH

Step 3: Determine moles of H₂SO₄ that react

According to the balanced equation, 2 moles of NaOH react with 1 mole of H₂SO₄

Therefore, the ratio giving moles of H₂SO₄ = Moles of NaOH ÷ 2

= 0.33 moles ÷ 2

= 0.165 moles

Step 4: Find the volume of H₂SO₄

Molarity indicates the concentration of the solution in moles per liter

Molarity = Moles ÷ Volume

By rearranging the formula, we find volume = Moles ÷ Molarity

= 0.165 moles ÷ 0.235 M

= 0.702 L

= 702 mL

Thus, the volume of the 0.235 M H₂SO₄ acid solution required equals 702 mL
8 0
2 months ago
The speed of light in a vacuum is 2.998 × 10 8 8 m/s. How long does it take for light to circumnavigate the Earth, which has a c
castortr0y [3046]

Answer:- 0.134 seconds

Solution:- The speed is given as 2.988*10^8\frac{m}{s} and the circumference is 24900 miles which is same as the distance light have to covered. It asks to calculate the time required to cover this distance by the light.

Unit conversion is needed from miles to meters since the speed is given in meters per second.

1 mile = 1609.34 meters

Thus, 24900mile(\frac{1609.34meter}{mile})

= 40072566 meters

Now, speed=\frac{distance}{time}

Rearranged for time, that gives: time=\frac{distance}{speed}

Inserting the values:

time=\frac{40072566meter}{2.988*10^8\frac{meter}{second}}

= 0.134 seconds

Hence, light would take 0.134 seconds to traverse the indicated distance. The answer without the unit is 0.134.

8 0
3 months ago
(58 g)/ (4L) reduce units to one
eduard [2782]
The result is 14.5 g L⁻¹. Here, the problem indicates to reduce the units to one. The existing units are g/L. To achieve a singular unit format, we can move L to the numerator, which can be executed as per the exponent laws; specifically, 1 / aˣ = a⁻ˣ. Thus, we can express 1 / L as L⁻¹. Consequently, the simplified unit remains g L⁻¹. However, remember to leave a space between two different units. This ultimately depicts a unit of density.
3 0
3 months ago
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