All categories

3 years ago

if 0.683 grams of silver chloride is produced how much (mass) silver nitrate would need to be reacted

1 answer:

4 0

The (mass) silver nitrate would needed to be reacted is calculated as follow

by assuming that AgNo3 reacted with Bacl

that is 2AgNo3 + BaCl2 ---> 2Agcl + Ba(No3)2

find the moles AgNo3 = 0.683g/169.87 g/mol =4.02 x10^-3 moles

by use of mole ratio between Agcl and Agno3 which is 2:2 this implies that the moles of AgNO3 also 4.02 x10^-3

mass= moles x molar mass

= (4.02 x10^-3) x 143.37= 0.576 grams

You might be interested in

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

. an ionic compound is composed of 34.95 g of iron and 15.05 g of oxygen. find the empirical formula of this compound.

Contact [7]

Empirical formula of ionic compound is FeO. In which the composition of atoms is 1 : 1.

Empirical formula of an ionic compound is defined as the formula which gives whole number ratio of atoms of various elements present in molecule of compund.

mass of iron in compound = 34.95 g

molar mass of iron = 55.8 g

mass of oxygen in compound = 15.05 g

molar mass of oxygen = 32 g

number of moles of iron present in the compound are ratio of mass of iron in compound/ molar mass of iron

number of moles of iron in compound= 34.95 / 55.8 = 0.6263 ~ 1

number of moles oxygen in compound= 15.05/ 32 = 0.473 ~ 0.5

the ratio of the number of oxygen atoms to number of iron atoms present in one formula unit of iron compund is 2×0.5 / 1 = 1 : 1

Hence , the required empirical formula of iron compound is FeO.

To learn more about Emiprical formula, refer:

brainly.com/question/1439914

#SPJ4

3 0

1 year ago

Describe the electrons in an atom of carbon in the ground state. Your response must

zepelin [54]

Based on the information given, it should be noted that the ground-state electron configuration of carbon is 1s2 2s2 2p2.

<h3>What is an electron?</h3>

Electrons are simply the subatomic particles which orbit the nucleus of an atom.

The arrangement of electrons in the atomic orbitals of an atom is known as the electron configuration. This can be determined by using a periodic table.

It should be noted that carbon is the sixth element with a total of 6 electrons in the periodic table. Thus, the atomic number Z = 6.

In conclusion, the ground-state electron configuration of carbon is 1s2 2s2 2p2.

Learn more about carbon on:

brainly.com/question/105003

8 0

2 years ago

Calculate the amount of energy , in Joules, required to raise the temperature of 15.5 g of liquid water from 0.00o C to 75.0 oC.

deff fn [24]

Answer:

10043.225 J

Explanation:

We'll begin by calculating the amount of heat needed to change ice to water since water at 0°C is ice. This is illustrated below:

Mass (m) = 15.5g

Latent heat of fussion of water (L) = 334J/g

Heat (Q1) =..?

Q1 = mL

Q1 = 15.5 x 334

Q1 = 5177 J

Next, we shall calculate the amount of heat needed to raise the temperature of water from 0°C to 75°C.

This is illustrated below:

Mass = 15.5g

Initial temperature (T1) = 0°C

Final temperature (T2) = 75°C

Change in temperature (ΔT) = T2 – T1 = 75 – 0 = 75°C

Specific heat capacity (C) of water = 4.186J/g°C

Heat (Q2) =?

Q2 = MCΔT

Q2 = 15.5 x 4.186 x 75

Q2 = 4866.225 J

The overall heat energy needed is given by:

QT = Q1 + Q2

QT = 5177 + 4866.225

QT = 10043.225 J

Therefore, the amount of energy required is 10043.225 J

8 0

3 years ago

Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---> CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu

Nitella [24]

ΔH = +438 kJ

We have three equations:

(I) N₂ + 3H₂ → 2NH₃; ΔH = -92 kJ

(II) H₂ +½O₂ → H₂O; ΔH = -286 kJ

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; ΔH = -632 kJ

From these, we must devise the target equation:

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = ?

_________________________________

The target equation has 2NH₃ on the left, so you reverse equation (I).

When you reverse an equation, you reverse the sign of its ΔH.

(V) 2NH₃ → N₂ + 3H₂; ΔH = +92 kJ

Equation (V) has 1N₂ on the right, and that is not in the target equation.

You need an equation with 1N₂ on the left.

Reverse Equation (III).

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ

Equation (VI) has ³/₂O₂ on the right, and that is not in the target equation.

You need ³/₂O₂ on the left.

Multiply Equation (II) by three.

When you multiply an equation by three, you multiply its ΔH by three.

(VII) 3H₂ +³/₂O₂ → 3H₂O; ΔH = -286 kJ

Now, you add equations (V), (VI), and (VII), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

_______________________________________

We get the target equation (IV):

(V) 2NH₃ → N₂ + 3H₂; ΔH = + 92 kJ

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ

(VII) 3H₂ +³/₂O₂ → 3H₂O; ΔH = -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = +438 kJ

7 0

3 years ago