All categories

3 years ago

Are the properties of Rubidium more similar to those of cesium or those of strontium?

Chemistry

1 answer:

fenix001 [56]3 years ago

5 0

More similar to Cesium

Explanation:

The properties of Rubidium are more similar to those of cesium compared to strontium.

Elements in the same group on the periodic table have similar chemical properties.

Rubidium and Cesium are located in the first group on the periodic table.
Other elements in this group are lithium, sodium, potassium and francium
Strontium belongs to the second group on the periodic table.
The first group have a ns¹ valence shells electronic configuration.
They are all referred to as alkali metals

Learn more:

Sodium brainly.com/question/6324347

Periodic table brainly.com/question/1704778

#learnwithBrainly

You might be interested in

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°

miskamm [114]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is -1835 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)$ $\Delta H_1=157kJ$ ( × 4)

(2) $P_4(g)+6Cl_2(g)\rightarrow 4PCl_3(g)$ $\Delta H_2=-1207kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(4\times (-157))+(1\times (-1207))=-1835kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -1835 kJ.

6 0

3 years ago

Four gases were combined in a gas cylinder with these partial pressures: 3.5 atm N2, 2.8 atm O2, 0.25 atm Ar, and 0.15 atm He.

Cerrena [4.2K]

Answer:

Answer :

The total pressure inside the cylinder is, 6.7 atm

The mole fraction of N_2 in the mixture is, 0.52

Solution :

First we have to calculate the total pressure inside the cylinder.

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.

P_T=p_{N_2}+p_{O_2}+p_{Ar}+p_{He}

Now put all the given values is expression, we get the total pressure inside the cylinder.

P_T=3.5+2.8+0.25+0.15=6.7atm

Now we have to calculate the mole fraction of N_2 in the mixture.

Formula used :

pN_2=XN_2 x P_T

where,

P_T = total pressure = 6.7 atm

pN_2 = partial pressure of nitrogen gas = 3.5 atm

XN_2 = mole fraction of nitrogen gas = ?

Now put all the given values in the above formula, we get

3.5atm=XN_2 x 6.7atm

XN_2=0.52

Therefore, the total pressure inside the cylinder is, 6.7 atm and the mole fraction of N_2 in the mixture is, 0.52

Hope it helps answer the question:)

5 0

3 years ago

Which of the following is the BEST description of our atmosphere?

tensa zangetsu [6.8K]

Answer:

I think the answer is B.

4 0

3 years ago

How many atoms is in h20

Nataly_w [17]

Answer:

3 atoms

Explanation:

2 hydrogen atoms and 1 oxygen atom

5 0

3 years ago

Read 2 more answers

Nitrogen has a half life of 10 days. If you have a 75 gram sample how much is remaining after 20 days ?

CaHeK987 [17]

There is 9.375 g of nitrogen. This is because when a half-life passes, the mass becomes 2 times less. Since two half-lives have passed, you need to divide by 4.

3 0

3 years ago