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Vera_Pavlovna [14]
2 years ago
7

Two groups complete an experiment requiring the mass measurement to be repeated three times in the experiment. Group I selects a

single member of the group to measure the mass. Group II allows any person of the group to measure the mass once, so they will be done more quickly. At the end of the experiment, the results for group I are closer to the actual than the results of group II. Explain why this might be.
Chemistry
1 answer:
Irina18 [472]2 years ago
3 0
Group 1 was more careful in choosing the person, whereas Group 2 might have been more careless. 
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A hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:
kupik [55]

hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

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En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.

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6 0
2 years ago
3. Where are gastropods found? Choose all the correct answers.
dalvyx [7]
D.

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4 0
3 years ago
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New questionWhat mass of calcium chloride (CaCl₂) would beproduced from the reaction of 125.9 g of hydrochloriacid (HCI) with ex
marshall27 [118]

Answer:

191.6 g of CaCl₂.

Explanation:

What is given?

Mass of HCl = 125.9 g.

Molar mass of CaCl₂ = 110.8 g/mol.

Molar mass of HCl = 36.4 g/mol.

Step-by-step solution:

First, we have to state the chemical equation. Ca(OH)₂ react with HCl to produce CaCl₂:

Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O.

Now, let's convert 125.9 g of HCl to moles using the given molar mass (remember that the molar mass of a compound can be found using the periodic table). The conversion will look like this:

125.9\text{ g HCl}\cdot\frac{1\text{ mol HCl}}{36.4\text{ g HCl}}=3.459\text{ moles HCl.}

Let's find how many moles of CaCl₂ are being produced by 3.459 moles of HCl. You can see in the chemical equation that 2 moles of HCl reacted with excess Ca(OH)₂ produces 1 mol of CaCl₂, so we state a rule of three and the calculation is:

3.459\text{ moles HCl}\cdot\frac{1\text{ mol CaCl}_2}{2\text{ moles HCl}}=1.729\text{ moles CaCl}_2.

The final step is to find the mass of CaCl₂ using the molar mass of CaCl₂. This conversion will look like this:

1.729\text{ moles CaCl}_2\cdot\frac{110.8\text{ g CaCl}_2}{1\text{ mol CaCl}_2}=191.6\text{ g CaCl}_2.

The answer would be that we're producing a mass of 191.6 g of CaCl₂.

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