In general, atomic radii: A. decrease down a group and remain constant across a period.
B. decrease down a group and increase across a period.
C. increase down a group and increase across a period.
D. increase down a group and remain constant across a period.
E. increase down a group and decrease across a period.

Answers

Answer 1

Answer:

Answer: The correct answer is Option E.

Explanation:

Atomic radius is defined as the total distance measured from the nucleus of an atom to the outermost shell.

Trend down the group:

Moving from top to bottom, a new shell gets add up around the nucleus and the outermost shell gets far away from the nucleus. Due to this, the distance between the nucleus and outermost shell increases, which results in the increase of atomic radii of the atom.

Trend across the period:

Moving from left to right in a period, more and more electrons gets add up in the same shell. The attraction between the last electron and the nucleus increases. This results in the shrinkage of size of an atom. Thus, decreasing the atomic radii of the atom.

Hence, the correct answer is Option E.

Answer 2

Answer:

Final answer:

E. increase down a group and decrease across a period. Atomic radii generally increase down a group due to extra electron shells and decrease across a period due to greater nuclear charge.

Explanation:

In general, the correct answer to this question is E: atomic radii increase down a group and decrease across a period on the Periodic Table. The atomic radii increase down a group due to the addition of extra electron shells. Each additional shell means a greater distance between the nucleus and the outermost electrons, which results in a larger atomic radius. On the other hand, as you move across a period from left to right, atomic radii typically decrease. This is due to an increase in positive charge in the nucleus which pulls the electrons closer, thus decreasing the atomic radius.

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Help me question 5 ASAP

Answers

I think it’s B.
I might be wrong, I’m just kinda going off based a little research and it says that it only happens in reptiles and teleost fish. The only option on there that’s related to it is the leatherback turtle.

A solution of NaOH has a concentration of 25.00% by mass. What mass of NaOH is present in 0.250 g of this solution? Use the periodic table in the toolbar if needed.

Answers

Answer : The mass of NaOH present in the solution is, 0.0625 grams

Explanation : Given,

Mass % = 25 %

Mass of solution = 0.250 g

Formula used :

$Mass\%=\frac{\text{Mass of}NaOH}{\text{Mass of solution}}* 100$

Now put all the given values in this formula, we get the mass of NaOH.

$25=\frac{\text{Mass of}NaOH}{0.250g}* 100$

$\text{Mass of}NaOH=0.0625g$

Therefore, the mass of NaOH present in the solution is, 0.0625 grams

The concentration of NaOH is 25.00% by mass, it means that 25.00% of the mass of the solution is of NaOH. Hence:

$m_(NaOH)=25.00\%* m_(solution)\Longrightarrow m_(NaOH)=(25)/(100)* 0.250~g\iff\n\n\boxed{m_(NaOH)=0.0625~g}$

HELP!!!!!!
I DON'T KNOW THE ORDER!!

Answers

Answer:

for 1 solid its freezing.

for 2 solid and liquid its melting

for 6 liquid to gas its evaporation and for 5 gas to liquid its condensation.

Explanation:

hope this helped :)

Answer:

solid->liquid= melting

liquid->solid= freezing

gas->liquid= consendation

liquid->gas= evaporation

The dry solute has a mass of 0.086 g. 0.113 g of water evaporated while heating. What is the concentration in G solute / mL solvent at 30.1 C? Remember: 1 g = 1 ml for waterA. 0.88
B. 0.0860
C. 0.113
D. 0.761

Answers

The dry solute has a mass of 0.086 g. 0.113 g of water evaporated while heating, the concentration in G solute / mL solvent at 30.1 C is 0.88.

To find the concentration in g solute / mL solvent, we first need to find the mass of the solvent left after evaporation:

Mass of water before evaporation = 0.113 g

Mass of water after evaporation = 0.113 g - 0.086 g = 0.027 g

Since 1 g of water has a volume of 1 mL, the volume of water after evaporation is also 0.027 mL.

Next, we need to convert the temperature to Kelvin:

T = 30.1 + 273.15 = 303.25 K

We can now use the formula:

concentration = (mass of solute / mass of solvent) / (1 - (mass of water evaporated / mass of solvent))

Plugging in the values we get:

concentration = (0.086 g / (0.027 g)) / (1 - (0.113 g / (0.086 g + 0.027 g)))

concentration = 0.88 g/mL

Therefore, the answer is A. 0.88.

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Answer:

Explanation:

worked for me on acellus

If the patient has to be administered a dosage of 2 tablets every 8 hours for 7 days,what is the number of tablets required for the prescribed dosage

Answers

42 tablets. 2*8*7=42

(II) To make a secure fit, rivets that are larger than the rivet hole are often used and the rivet is cooled (usually in dry ice) before it is placed in the hole. A steel rivet 1.872cm in diameter is to be placed in a hole 1.870cm in diameter in a metal at 22°C. To what temperature must the rivet be cooled if it is to fit in the hole

Answers

Given:

Rivet diameter, $d_(r)$ = 1.872 cm

Hole diameter, $d_(h)$ = 1.870 cm

Temperature, $T_(2)$ = 22 °C

Formula Used:

$\alpha = (\Delta d)/(d* \Delta T)$

where,

$\alpha$ = coefficient of linear expansion

$\Delta d$ = change in diameter = $d_(h) - d_(r)$

$\Delta T}$ = change in temperature = $T_(2) - T_(1)$

Solution:

we know that coefficient of linear expansion of steel, $\alpha$ = $12* 10^(-6)/^(\circ)C$

Using the above formula :

$\alpha = (\Delta d)/(d* \Delta T)$

$12* 10^(-6)/^(\circ)C$ = \frac{1.870 - 1.872}{1.872\times \T_{2} - T_{1}}[/tex]

$T_(2) - 20/^(\circ)C$ = \frac{1.870 - 1.872}{12\times 10^{-6}}}[/tex]

$T_(2) = -67.03/^(\circ)C$

Therefore, the rivet must be cooled to $-67.03/^(\circ)C$

Final answer:

The question involves the concept of thermal expansion in Physics. By knowing the initial diameter of the rivet and hole, as well as the ambient temperature, we can use the thermal expansion formula to calculate the temperature to which the steel rivet must be cooled to fit into the hole.

Explanation:

The subject in question pertains to Physics and specifically to the concept of thermal expansion. This indicates how objects (in this case, a steel rivet) tend to change in volume or shape as a response to a change in temperature. The diameter of the rivet when cooled will decrease slightly, allowing it to fit into the smaller hole.

To find the temperature to which the rivet needs to be cooled, we require knowledge of the thermal expansion coefficient of steel, which (for generalization) can be averaged to around 0.000012 (1/°C). The formula to calculate the change in diameter (Δd) is:

Δd = α * d * ΔT

where α is the coefficient of linear expansion, d is the original diameter, and ΔT is the change in temperature. Knowing the initial diameter of the rivet and the hole it must fit into, together with the ambient temperature (22°C), we can rearrange this formula to find the cooling temperature needed for the rivet to fit into the hole.

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