How many bonding pairs in br2o

15.3 g of nano3 were dissolved in 100g of water in a calorimeter. The temperature of the water drops from 25.00°c to 21.56°c. Ca

Arturiano [62]

Answer:

0.259 kJ/mol ≅ 0.26 kJ/mol.

Explanation:

To solve this problem, we can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat absorbed by ice (Q = ??? J).

m is the mass of the ice (m = 100.0 g).

c is the specific heat of water (c of ice = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 21.56°C - 25.0°C = -3.44°C).

∵ Q = m.c.ΔT

∴ Q = (100.0 g)(4.186 J/g.°C)(-3.44°C) = -1440 J = -1.44 kJ.

∵ ΔH = Q/n

n = mass/molar mass = (100.0 g)/(18.0 g/mol) = 5.556 mol.

∴ ΔH = (-1.44 kJ)/(5.556 mol) = 0.259 kJ/mol ≅ 0.26 kJ/mol.

3 0

3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago

which of the elements below has a smaller atomic radius. element1: 1s2 2s2 2p4 element2: 1s2 2s2 2p5

stiks02 [169]

Answer:

both are same give different number

8 0

2 years ago

6. What is a polyatomic ion? (You should also memorize the name, formula, and charge of the polyatomic ions listed in the lectur

strojnjashka [21]

A polyatomic ion is a group of elements that has a charge that is not 0. for example the charge of P04 is -3

6 0

3 years ago

Write the formula of the coordination compound pentaamminecarbonatocobalt(III) iodide. Enclose the coordination complex in squar

7nadin3 [17]

Answer:

[Co(NH3)5CO3]I3

Explanation:

The naming of coordination compounds follows certain rules specified by IUPAC. Usually, the name of the complex makes it quite easy to deduce its structure.

"Pentaamine" means that there are five NH3 ligands as shown in the structure. The ligand carbonato is CO3^2-. It has no prefix attached to it in the IUPAC name of the complex hence there is only one carbonato ligand present(recall that the complex has a coordination number of six). I did not enclose it within parenthesis as required in the question.

Lastly the III that appeared after the metal name "cobalt" shows its oxidation state. The iodide counter ions must then be 3 in number in order to satisfy this primary valency of the metal hence the inclusion of I3 in the structure of the complex.

6 0

3 years ago