All categories

3 years ago

Which is one way scientists indicate how precise And accurate their experimental measurements are?

Chemistry

1 answer:

nadezda [96]3 years ago

8 0

<span>. They make sure their experiments can be repeated</span>

You might be interested in

In Universe , recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quant

lawyer [7]

Answer:

According to Hund's rule and the Aufbau principle in which the orbitals must be filled with electrons, they are not strictly applied in the real universe, because the intermediate and electron-filled atomic orbitals are very stable . Because there are four d-orbitals in universe L, a typical half-full configuration will be xd4 and its full configuration will be xd8, where x is the primary orbital for any specific element. Here is an example:

Vahadium ₂₃V

in real universe: [Ar]₈ 3d³4s²

in universe L: [Ar]₁₈ 3d⁴4s¹

Chromium

in real universe: [Ar]₈ 3d⁵4s¹

in universe L: [Ar]₁₈ 3d⁴4s²

Explanation:

8 0

3 years ago

If 9.8g water is used in electrolysis, what is the percent yield if 5.6g of oxygen was

ANEK [815]

Answer:

Approximately $64\%$ .

Explanation:

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

The actual yield of $\rm O_2$ was given. The theoretical yield needs to be calculated from the quantity of the reactant.

Balance the equation for the hydrolysis of water:

$\rm 2\, H_2O \, (l) \to 2\, H_2\, (g) + O_2\, (g)$ .

Note the ratio between the coefficient of $\rm H_2O\, (g)$ and $\rm O_2\, (g)$ :

$\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ .

This ratio will be useful for finding the theoretical yield of $\rm O_2\, (g)$ .

Look up the relative atomic mass of hydrogen and oxygen on a modern periodic table.

$\rm H$ : $1.008$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm H_2O$ and $\rm O_2$ :

$M(\mathrm{H_2O}) =2\times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{O_2}) =2\times 15.999 = 31.998\; \rm g \cdot mol^{-1}$ .

Calculate the number of moles of molecules in $9.8\; \rm g$ of $\rm H_2O$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{9.8\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \approx 0.543991\;\rm g \cdot mol^{-1}$ .

Make use of the ratio $\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ to find the theoretical yield of $\rm O_2$ (in terms of number of moles of molecules.)

$\begin{aligned} n(\mathrm{O_2}) &= \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} \cdot n(\mathrm{H_2O}) \\ &\approx \frac{1}{2} \times 0.543991\; \rm mol \approx 0.271996\; \rm mol \end{aligned}$ .

Calculate the mass of that approximately $0.271996\; \rm mol$ of $\rm O_2$ (theoretical yield.)

$\begin{aligned}m(\mathrm{O_2}) &= n(\mathrm{O_2}) \cdot M(\mathrm{O_2}) \\ &\approx 0.271996\; \rm mol \times 31.998\; \rm g \cdot mol^{-1} \approx 8.70331 \; \rm g \end{aligned}$ .

That would correspond to the theoretical yield of $\rm O_2$ (in term of the mass of the product.)

Given that the actual yield is $5.6\; \rm g$ , calculate the percentage yield:

$\begin{aligned}\text{Percentage Yield} &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \\ &\approx \frac{5.6\; \rm g}{8.70331\; \rm g} \times 100\% \approx 64\%\end{aligned}$ .

4 0

2 years ago

Can someone please help me with my science research I kind of need help in a few things that I listed, I’m currently working on

dalvyx [7]

Answer:

For your first question, Curium does not occur naturally on Earth, meaning that it is not produced naturally on Earth. However, it can be formed in nuclear reactors.

For your second question, Curium has been used to provide power to electrical equipment used on space missions, but doesn't seem to be that important overall.

Explanation:

Hope this helped!

7 0

2 years ago

Which of the following pure liquids is the best solvent for sodium fluoride?a. CCl4b. C2C6c. HCld. BCl3e. PCl5

Julli [10]

Answer:

HCl

Explanation:

The best solvent for NaF is a polar liquid. The only liquid having a significant dipole moment among the options is HCl due to the large electro negativity difference between hydrogen and chlorine.

The polar solvent can interact with the NaF via its dipoles such that the NaF dissolves due to ion-dipole interaction.

4 0

3 years ago

Why does lithium form only lithium oxide and not peroxide or superoxide

Alexandra [31]

Answer:

Lithium does form a peroxide as well as an oxide on burning in air and I suspect the low temperature reaction with air forms a significant amount of peroxide.

8 0

3 years ago