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3 years ago

In whare atoms in group 1 and atoms in group 18 different?

Chemistry

1 answer:

sdas [7]3 years ago

8 0

Group 1 contains metals while group 18 contains noble gases.

So group 1 is different from group 18 as they both contains different types of atoms as group 1 contains metals while group 18 contains noble gases.

There is 1 valence shell electron in group 1 , they are highly reactive while valence shell of group 18 is fully filled and they are least reactive .

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A 40.0 L balloon is filled with air at sea level (1 atm @ 25 oC). It is then tied to a rock and thrown into a cold lake and it s

velikii [3]

To solve this question you need to calculate the number of the gas molecule. The calculation would be:
PV=nRT
n=PV/RT
n= 1 atm * 40 L/ (0.082 L atm mol-1K-1 * 298.15K)
n= 1.636 moles

The volume at bottom of the lake would be:
PV=nRT
V= nRT/P
V= (1.636 mol * 277.15K* 0.082 L atm mol-1K-1 )/ 11 atm= 3.38 L

8 0

3 years ago

Which two atoms could be the "before'' and ''after'' of an alpha ejection

spin [16.1K]

The answer would be uranium and thorium. When an alpha ejects a particle, it will create a new atom. So, when uranium ejects an alpha particle, it will produce thorium. They call this process as the alpha decay. Alpha decay often happens on atoms that are abundant nuclei such as uranium, radium, and thorium.

4 0

4 years ago

Since helium is lighter than air, it is difficult to measure the mass of a sample, so to find out the mass of helium in a birthd

d1i1m1o1n [39]

Answer:

0.7457 g is the mass of the helium gas.

Explanation:

Given:

Pressure = 3.04 atm

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (25.0 + 273.15) K = 298.15 K

Volume = 1.50 L

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

3.04 atm × 1.50 L = n × 0.0821 L.atm/K.mol × 298.15 K

⇒n = 0.1863 moles

Molar mass of helium = 4.0026 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.1863\ mole= \frac{Mass}{4.0026\ g/mol}$

$Mass_{He}= 0.7457\ g$

0.7457 g is the mass of the helium gas.

6 0

3 years ago

N2H4 + N2O4 --> N2 + H2O

ahrayia [7]

Answer:

The limiting reagent is N2O4
14,09g

Explanation:

First, we adjust the reaction.

$2N_{2} H_{4}$ + $N_{2} O_{4}$ ⇄ $6N_{2} + 4H_{2}O$

Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using $N_{2} H_{4}$ to form $H_{2}O$

$molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4} }{32,04\frac{g}{mol} N_{2} H_{4} }$

$molH_{2} O = 1, 125 mol$

Using $N_{2} O_{4}$ to form $H_{2} O$

$molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4} }{92\frac{g}{mol} N_{2} O_{4} }$