As the atoms of the elements from atomic

Chemistry

1 answer:

In-s [12.5K]2 years ago

7 0

Answer:

smaller, greater

Explanation:

Moving from left to right in the periodic table, we notice that the magnitude of nuclear charge increases with increasing atomic number.

Recall that across the period, nuclear charge increases without a corresponding increase in the number of shells hence the size of the atom (atomic radius) contracts due to a greater pull of the nuclear charge on the electrons.

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A jeweler guarantees that a piece of jewelry is at least 95% gold, by mass. You consider buying a piece of gold jewelry that wei

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The density of the sample is:
Density = mass / volume
Density = 9.85 / 0.675
Density = 14.6 g/cm³

If the sample has 95% gold, and 5% silver, its density should be:
0.95 x 19.3 + 0.05 x 10.5
Theoretical density = 18.9 g/cm³

The difference in theoretical and actual densities is very large, making it likely that the jeweler was not telling the truth.

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Which factor has the least effect on the rate of solution of a solid in a liquid?

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The pressure will not affect the rate of solution.

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Rina8888 [55]

B mass but not of charge

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What is chemical weathering of rock

balu736 [363]

Chemical weathering is the weakening and subsequent disintegration of rock by chemical reactions.

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If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution

Helen [10]

This is a straightforward dilution calculation that can be done using the equation

$M_1V_1=M_2V_2$

where M₁ and M₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and V₁ and V₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (M₁) and the initial (V₁) and final (V₂) volumes, and we want to find the final concentration (M₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for M₂:

$M_2=\frac{M_1V_1}{V_2}.$

Substituting in our values, we get

$\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].$

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

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