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LUCKY_DIMON [66]
2 years ago
9

which of the following testable questions will provide evidence that elements in the same group have similar properties?

Chemistry
1 answer:
Usimov [2.4K]2 years ago
4 0

The testable question which will provide evidence that elements in the same group have similar properties is valency

<h3>What is the valency?</h3>

In essence, how many electrons are present in their outermost shell.

Valency and Groups in the periodic table, elements are arranged in order of their atomic numbers and hence, the periodic table is a systematic arrangement of elements in an array of vertical columns called Groups and horizontal arrays called Periods.

  • Complete question here:

which of the following testable questions will provide evidence that elements in the same group have similar properties A. Valency B. Orbit C. Group D. Period

Since, the reactive capacities of elements is dependent on the number of electrons on its outermost shell, we can conclude on this note that, Elements in the same group have similar properties.

Learn more on Groups and Valency:

brainly.com/question/1645905

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sasho [114]
The balanced equation for the above reaction is 
2K₃PO₄   + 3NiCl₂  ---> 6KCl + Ni₃(PO₄)₂
stoichiometry of K₃PO₄  to NiCl₂ is 2:3
the number of NiCl₂ moles reacted - 0.0110 mol/L x 0.154 L = 1.69 x 10⁻³ mol
if 3 mol of NiCl₂ reacts with - 2 mol of K₃PO₄ 
then 1.69 x 10⁻³ mol of NiCl₂ reacts with - 2/3 x 1.69 x 10⁻³  = 1.13 x 10⁻³ mol of K₃PO₄
molarity of K₃PO₄ solution given - 0.205 M
there are 0.205 mol in 1 L
therefore 1.13 x 10⁻³ mol are in - 1.13 x 10⁻³ mol / 0.205 mol/L = 5.51 mL
volume of K₃PO₄ required - 5.51 mL
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What is anything that has mass and volume
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3 years ago
What is the de Broglie wavelength (in meters) of a 45-g golf ball traveling at 72 m/s?
Cerrena [4.2K]

Answer: The de broglie wavelength is 2.037 \times 10^{-34} m.

Explanation:

Calculate  \lambda = \frac{h}{p}as follows.

          \lambda = \frac{h}{p}

where,

          h = plank's constant = 6.6 \times 10^{-34} m^{2} kg/s

         p = momentum = mass \times velocity

Putting the values in the formula as follows.

        \lambda = \frac{h}{mass \times velocity}

                               =  \frac {6.6 \times 10^{-34} m^{2} kg/s}{0.045 kg \times 72 m/s}                        

                               =  2.037 \times 10^{-34} m

Thus, the de broglie wavelength is 2.037 \times 10^{-34} m.

                               

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3 years ago
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