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3 years ago

Calculate the grams of so2 gas present at stp in a 5.9 l container.

Chemistry

1 answer:

nata0808 [166]3 years ago

7 0

Answer: The mass of sulfur dioxide gas at STP for given amount is 16.8 g

Explanation:

At STP conditions:

22.4 L of volume is occupied by 1 mole of a gas.

So, 5.9 L of volume will be occupied by = $\frac{1mol}{22.4L}\times 5.9L=0.263mol$

Now, to calculate the mass of a substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of sulfur dioxide gas = 0.263 mol

Molar mass of sulfur dioxide gas = 64 g/mol

Putting values in above equation, we get:

$0.263mol=\frac{\text{Mass of sulfur dioxide gas}}{64g/mol}\\\\\text{Mass of sulfur dioxide gas}=(0.263mol\times 64g/mol)=16.8g$

Hence, the mass of sulfur dioxide gas at STP for given amount is 16.8 g

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A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.

BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

$\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\ &=185.59\text{ g/mol} \end{aligned}$

Dimensional Analysis:

$\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }$

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

8 0

3 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

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

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

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

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

3 years ago

Under standard-state conditions, which of the following species is the best reducing agent? a. Ag+ b. Pb c. H2 d. Ag e. Mg2+

eimsori [14]

Answer: The correct answer is Option b.

Explanation:

Reducing agents are defined as the agents which help the other substance to get reduced and itself gets oxidized. They undergo oxidation reaction.

$X\rightarrow X^{n+}+ne^-$

For determination of reducing agents, we will look at the oxidation potentials of the substance. Oxidation potentials can be determined by reversing the standard reduction potentials.

For the given options:

Option a: $Ag^+$

This ion cannot be further oxidized because +1 is the most stable oxidation state of silver.

Option b: $Pb$

This metal can easily get oxidized to $Pb^{2+}$ ion and the standard oxidation potential for this is 0.13 V

$Pb\rightarrow Pb^{2+}+2e^-;E^o_{(Pb/Pb^{2+})}=+0.13V$

Option c: $H_2$

This metal can easily get oxidized to $H^{+}$ ion and the standard oxidation potential for this is 0.0 V

$H_2\rightarrow 2H^++2e^-;E^o_{(H_2/H^{+})}=0.0V$

Option d: $Ag$

This metal can easily get oxidized to $Ag^{+}$ ion and the standard oxidation potential for this is -0.80 V

$Ag\rightarrow Ag^{+}+e^-;E^o_{(Ag/Ag^{+})}=-0.80V$

Option e: $Mg^{2+}$

This ion cannot be further oxidized because +2 is the most stable oxidation state of magnesium.

By looking at the standard oxidation potential of the substances, the substance having highest positive $E^o$ potential will always get oxidized and will undergo oxidation reaction. Thus, considered as strong reducing agent.

From the above values, the correct answer is Option b.

8 0

3 years ago

Identify the Brønsted-Lowry acid, the Brønsted-Lowry base, the conjugate acid, and the conjugate base in each reaction:

ExtremeBDS [4]

Answer:

C5H5N is the base and C5H5NH+ is the conjugate acid

H2O is the acid and OH− is the conjugate base

Explanation:

Hydrogen + is also called a proton

C5H5N is the base because it receives the proton (H+) and C5H5NH+ is its conjugate acid

H2O is the acid because it gives up the proton and OH− is the conjugate base because it is capable of receiving the proton

Answer:

HNO3 is the acid and NO3- is the conjugate base

H2O is the base and H3O+ is the conjugate acid

Explanation

HNO3 is the acid and NO3− is its conjugate base, capable of receiving a proton

H2O is the base because it receives the proton and H3O+ is a conjugate acid capable of giving up the proton.

3 0

3 years ago

Which part of an of the helium atom is positively charged

olchik [2.2K]

Nucleus because it is made of protons (which means positively charged)

6 0

3 years ago