Find the radius

Consider the following reaction of calcium hydride (CaH2) with molten sodium metal: CaH2(s) + 2 Na(l) -> 2 NaH(s) + Ca(l) Ide

vodka [1.7K]

Answer:

This is not a redox reaction. None of the species are reduced and none are oxidized

Explanation:

In a redox reaction at least one species must be oxidized and another reduced. You see this by a change in oxidation number. In this question the oxidation numbers are the same before and after the reaction.

Ca 2

H -1

Na +1

I -1

4 0

4 years ago

The average molecular speed in a sample of Ar gas at a certain temperature is 391 m/s. The average molecular speed in a sample o

adoni [48]

Answer:

550 m/s

Explanation:

The average molecular speed (v) is the speed associated with a group of molecules on average. We can calculate it using the following expression.

$v = \sqrt{\frac{3 \times R \times T}{M} }$

where,

R: ideal gas constant

T: absolute temperature

M: molar mass of the gas

We can use the info of argon to calculate the temperature for both samples.

$T = \frac{v^{2} \times M}{3 \times R} = \frac{(391m/s)^{2} \times 39.95g/mol}{3 \times 8.314J/k.mol} = 2.45 \times 10^{5} K$

Now, we can use the same expression to find the average molecular speed in a sample of Ne gas.

$v = \sqrt{\frac{3 \times R \times T}{M} } = \sqrt{\frac{3 \times (8.314J/k.mol) \times 2.45 \times 10^{5}K }{20.18g/mol} } = 550 m/s$

7 0

3 years ago

What is the mass in grams of each of the following samples? 1.505 mol of Ti 0.337 mol of Na

Masja [62]

Answer:

72.04g Ti and 7.75g Na

Explanation:

1.505 mol Ti x (47.87g/1 mol) = 72.04g Ti

0.337 mol Na x (22.99g/1 mol) = 7.75g Na

5 0

3 years ago

Consider the solubilities of a particular solute at two different temperatures. Temperature ( ∘ C ) Solubility ( g / 100 g H 2 O

grin007 [14]

Answer:

21.28 grams solute can be added if the temperature is increased to 30.0°C.

Explanation:

Solubility of solute at 20°C = 32.2 g/100 grams of water

Solute soluble in 1 gram of water = $\frac{32.2}{100}g=0.322 g$

Mass of solute in soluble in 56.0 grams of water:

$0.322\times 56.0=18.032 g$

Solubility of solute at 30°C = 70.2g/100 grams of water

Solute soluble in 1 gram of water = $\frac{70.2}{100}g=0.702 g$

Mass of solute in soluble in 56.0 grams of water:

$0.702 \times 56.0=39.312 g$

If the temperature of saturated solution of this solute using 56.0 g of water at 20.0 °C raised to 30.0°C

Mass of solute in soluble in 56.0 grams of water 20.0°C = 18.032 g

Mass of solute in soluble in 56.0 grams of water at 30.0°C = 39.312 g

Mass of of solute added If the temperature of the saturated solution increased to 30.0°C:

39.312 g - 18.032 g = 21.28 g

21.28 grams solute can be added if the temperature is increased to 30.0°C.

8 0

3 years ago

Pentaborane−9 (B5H9) is a colorless, highly reactive liquid that will burst into flames when exposed to oxygen. The reaction is

mote1985 [20]

Answer : The heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

$2B_5H_9(l)+12O_2(g)\rightleftharpoons 5B_2O_3(s)+9H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(n_{(B_2O_3)}\times \Delta H^o_f_{(B_2O_3)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(B_5H_9)}\times \Delta H^o_f_{(B_5H_9)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol$