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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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What happens when you combine ammonia and formaldehyde?

bija089 [108]

Answer:

The mechanism for the formation of hexamethylenetetraamine predicts the formation of aminomethanol from the addition of ammonia to formaldehyde. This molecule subsequently undergoes unimolecular decomposition to form methanimine and water.

Explanation:

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2 years ago

What does multivalent mean???

salantis [7]

In the context of multivalent ions, it is when it has multiple oxidative states.

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2 years ago

A gas contains 75.0 wt% methane, 10.0% ethane, 5.0% ethylene, and the balance water. (a) Calculate the molar composition of this

NeTakaya

Answer:

a) molar composition of this gas on both a wet and a dry basis are

5.76 moles and 5.20 moles respectively.

Ratio of moles of water to the moles of dry gas =0.108 moles

b) Total air required = 68.51 kmoles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

Explanation:

Let assume we have 100 g of mixture of gas:

Given that :

Mass of methane =75 g

Mass of ethane = 10 g

Mass of ethylene = 5 g

∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)

Their molar composition can be calculated as follows:

Molar mass of methane $CH_4}= 16 g/mol$

Molar mass of ethane $C_2H_6= 30 g/mol$

Molar mass of ethylene $C_2H_4 = 28 g/mol$

Molar mass of water $H_2O=18g/mol$

number of moles = $\frac{mass}{molar mass}$

Their molar composition can be calculated as follows:

$n_{CH_4}= \frac{75}{16}$

$n_{CH_4}=$ 4.69 moles

$n_{C_2H_6} = \frac{10}{30}$

$n_{C_2H_6} =$ 0.33 moles

$n_{C_2H_4} = \frac{5}{28}$

$n_{C_2H_4} =$ 0.18 moles

$n_{H_2O}= \frac{10}{18}$

$n_{H_2O}=$ 0.56 moles

Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles

= 5.76 moles

Total moles of gas for dry basis = (5.76 - 0.56)moles

= 5.20 moles

Ratio of moles of water to the moles of dry gas = $\frac{n_{H_2O}}{n_{drygas}}$

= $\frac{0.56}{5.2}$

= 0.108 moles

b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:

$CH_4 + 2O_2_{(g)} ------> CO_2_{(g)} +2H_2O$

4.69 2× 4.69

moles moles

$C_2H_6+ \frac{7}{2}O_2_{(g)} ------> 2CO_2_{(g)} + 3H_2O$

0.33 3.5 × 0.33

moles moles

$C_2H_4+3O_2_{(g)} ----->2CO_2+2H_2O$

0.18 3× 0.18

moles moles

Mass flow rate = 100 kg/h

Their Molar Flow rate is as follows;

$CH_4 = 4.69 k moles/h\\C_2H_6 = 0.33 k moles/h\\C_2H_4=0.18kmoles/h$

Total moles of $O_2$ required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles

= 11.075 k moles.

In 1 mole air = 0.21 moles $O_2$

Thus, moles of air required = $\frac{1}{0.21}*11.075$

= 52.7 k mole

30% excess air = 0.3 × 52.7 k moles

= 15.81 k moles

Total air required = (52.7 + 15.81 ) k moles/h

= 68.51 k moles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

5 0

3 years ago

Acetic acid is a weak acid with a pKa of 4.76. What is the concentration of acetate in a buffer solution of 0.2M at pH 4.9. Give

Ghella [55]

Answer:

$[base]=0.28M$

Explanation:

Hello,

In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:

$pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\$

$[base]=1.38[acid]=1.38*0.20M=0.28M$

Regards.

6 0

2 years ago

A standard solution contained 0.8 mg/mL. A student took 2 mL of the standard solution and added 10 mL of water. What is the new/

galben [10]

To get the concentration of the second solution let us use the following formulae

C1V1=C2V2 where C1 is concentration of first solution and V1 is the volume of solution first solution. on the other hand C2 is the concentration of second solution and V2 is the volume of second solution.

therefore

0.8×2=(2+10)×C2
1.6 =12×C2
1.6/12=C2
C2 = 0.1333mg/mL

7 0

3 years ago

Read 2 more answers