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Inessa05 [86]
3 years ago
10

How are elements with atomic masses greater than Uranium’s made?

Chemistry
1 answer:
JulsSmile [24]3 years ago
7 0
Atoms heavier that uranium do not exist in nature.they must be synthesized in a particle accelerator.
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HELP!
Elza [17]

Answer:

  • <u><em>It will be less than 26 °C as water has a relatively higher specific heat than sand.</em></u>

Explanation:

The <em>specific heat </em>of a substance is the amount of heat energy absorbed by one unit of mass of the substance when its temperature increases one unit.

From that, you can derive the equation for the specific heat of a substance:

  • specific heat = heat / (mass × ΔT)

Thus, assuming that all the heat provided by the lamp to both samples is the same and, as given, the amount (mass) of both samples is also the same, you have that the specific heat of the samples will be:

  • specific heat = constant / ΔT

So, specific heat and ΔT are inversely related.

It is known that water has a higher specific heat than sand (that is why the sand on the shore of a beach is, during the day, hotter than the water and your feet get burned when you walk on a sandy beach on a sunny day).

Then, since the specific heat of water is greater than the specific heat of sand, the increase of temperature of water will be lower and, consequently, water will reach a lower final temperature than sand, when equal amounts of water and sand are heated as described in the experiment. This is the second choice: the final temperature of water is less than 26°C as water has a relatively higher specific heat than water.

6 0
2 years ago
How many liters of a 3 M NaOH stock solution would you need to make 552 mL of a 105 mM NaOH dilution
Ganezh [65]

Answer:

0.01932 L

Explanation:

First we <u>convert 105 mM to M</u>:

  • 105 mM / 1000 = 0.105 M

Next we <u>convert 552 mL to L</u>:

  • 552 mL / 1000 = 0.552 L

Then we use the following equation:

  • C₁V₁=C₂V₂

Where:

  • C₁ = 3 M
  • V₁ = ?
  • C₂ = 0.105 M
  • V₂ = 0.552 L

We<u> input the given data</u>:

  • 3 M * V₁ = 0.105 M * 0.552 L

And <u>solve for V₁</u>:

  • V₁ = 0.01932 L
7 0
3 years ago
Put the list in chronological order (1–5).
Leokris [45]

Explanation:

Filtration is a separation technique in which solid particles suspended in liquid medium are separated by allowing the mixture through the pores of the filter paper. By this solid particles get collect on filter paper and liquid drains out from the pores of the filter paper.

The chronological order for given steps will be:

  1. Weigh and fold the filter paper.
  2. Place the filter paper in the funnel, then place the funnel in the Erlenmeyer flask.
  3. Allow the solid/liquid mixture to drain through the filter.
  4. Use water to rinse the filter paper containing the mixture.
  5. Weigh the dried filter paper and copper.
4 0
3 years ago
Read 2 more answers
Oxidation number of au in kaucl4
DanielleElmas [232]
KauCl4 :

K = + 1 

au = + 7

Cl = - 2

hope this helps!

3 0
3 years ago
Read 2 more answers
The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹  

Explanation:

Assume the rate law is  

rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

8 0
2 years ago
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