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3 years ago

An increase in the atomic number________the atomic radius moving from left to right across a period.

Chemistry

1 answer:

frez [133]3 years ago

4 0

Answer:

decrease

Explanation:

Atomic radius :

It is the distance from the center of nucleus to the outer most electronic shell.

Trend along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required

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3. What is the mass of 5 moles of Hydrogen sulfate?I

solong [7]

Answer:

490

98 for 1 mole, Hence for 5 moles 5 X 98 =490.

Explanation:

Brainliest please?

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3 years ago

Consider the followong balanced reaction. What mads in g of co2 can be formed from 288 mg of o2. Assume that there is excess c3h

Gwar [14]

Answer:

0.264 g of $CO_2$ can be formed from 288 mg of $O_2$

Explanation:

The balanced chemical equation is

$2 C_3 H_7 OH+9 O_2> 6 CO_2+8 H_2 O$

The conversions are

Mass in mg $O_2$ is converted to mass in g $O_2$

Mass in g $O_2$ is converted to moles $O_2$ by dividing with molar mass

Moles $O_2$ is converted to moles $CO_2$ by using the mole ratio of $O_2:CO_2$ is 9 : 6

Moles $CO_2$ is converted to mass $CO_2$ by multiplying with molar mass $CO_2$

mass in mg $O_2$ > mass in g $O_2$ >moles $O_2$ > moles $CO_2$ > mass $CO_2$

$288mg O_2 \times \frac{(1g O_2)}{(1000mg O_2 )} \times \frac {(1molO_2)}{(32gO_2 )}\times\frac {(6mol CO_2)}{(9mol O_2 )} \times \frac {(44.0 gCO_2)}{(1mol CO_2 )}$

=0.264g (Answer)

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3 years ago

A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the

horrorfan [7]

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution : Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5 $Cm^{3}$

First, find the Density of given substance.

Formula used :

$Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}$

Now,put all the values in this formula, we get

$Density=\frac{46.9 g}{3.5 Cm^{3} }$ = 13.4 g/ $Cm^{3}$

So, we conclude that the density of given substance (13.4 g/ $Cm^{3}$ ) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ $Cm^{3}$ respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

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