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3 years ago

Write the molecular, ionic, and net ionic equations for the reaction of Sr(NO3)2(aq) and Li2SO4(aq).

Chemistry

1 answer:

denis23 [38]3 years ago

6 0

Answer:

Molecular: $Sr(NO_3)_2(aq) +Li_2SO_4(aq)\rightarrow SrSO_4(s)+2LiNO_3(aq)$

Ionic: $Sr^{2+}(aq)+2NO_3^-(aq) +2Li^{2+}(aq)+SO_4(aq)^{2-}\rightarrow SrSO_4(s)+2Li^+(aq)+2NO_3^-(aq)$ Net ionic: $Sr^{2}(aq)+SO_4(aq)^{2-}\rightarrow SrSO_4(s)$

Explanation:

Hello!

In this case, since the molecular, ionic and net ionic equations show the complete molecules, ions and resulting ions respectively, for the reaction between strontium nitrate and lithium sulfate, we can notice the formation of solid strontium sulfate and lithium nitrate as shown below:

$Sr(NO_3)_2(aq) +Li_2SO_4(aq)\rightarrow SrSO_4(s)+2LiNO_3(aq)$

Which is the molecular equation showing both reactants and products as molecules. Then, the ionic equation shows all the reactants and products as ions, considering that aqueous solutions dissociate whereas solid, liquid and gaseous molecules do not, therefore, we obtain:

$Sr^{2+}(aq)+2NO_3^-(aq) +2Li^{2+}(aq)+SO_4(aq)^{2-}\rightarrow SrSO_4(s)+2Li^+(aq)+2NO_3^-(aq)$

Finally, for the net ionic equation, we cancel out the spectator ions, which are those at both reactants and products:

$Sr^{2+}(aq)+SO_4(aq)^{2-}\rightarrow SrSO_4(s)$

Best regards!

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The product gas is then passed through a concentrated solution of KOH to remove the CO2. After passage through the KOH solution,

Kamila [148]

<u>Answer:</u> The mass percent of nitrogen gas in the compound is 13.3 %

<u>Explanation:</u>

Assuming the chemical equation of the compound forming product gases is:

$\text{Compound}\xrightarrow[CuO(s)]{Hot}N_2(g)+CO_2(g)+H_2O(g)$

Now, the product gases are treated with KOH to remove carbon dioxide.

We are given:

$p_{Total}=726torr\\P_{water}=23.8torr\\$

So, pressure of nitrogen gas will be = $p_{Total}-p_{water}=726-23.8=702.2torr$

To calculate the number of moles of nitrogen, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of nitrogen gas = 702.2 torr = 0.924 atm (Conversion factor: 1 atm = 760 torr)

V = Volume of nitrogen gas = 31.8 mL = 0.0318 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of nitrogen gas = $25^oC=[25+273]K=298K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

$0.924atm\times 0.0318L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\n_{mix}=\frac{0.924\times 0.0318}{0.0821\times 298}=0.0012mol$

To calculate the mass of nitrogen gas, we use the equation:

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

Molar mass of nitrogen gas = 28 g/mol

Moles of nitrogen gas = 0.0012 moles

Putting values in above equation, we get:

$0.0012mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.0012mol\times 28g/mol)=0.0336g$

To calculate the mass percent of nitrogen gas in compound, we use the equation:

$\text{Mass percent of nitrogen gas}=\frac{\text{Mass of nitrogen gas}}{\text{Mass of compound}}\times 100$

Mass of compound = 0.253 g

Mass of nitrogen gas = 0.0336 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{0.0336g}{0.253g}\times 100=13.3\%$

Hence, the mass percent of nitrogen gas in the compound is 13.3 %

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3 years ago

Eq-43 which type of fire extinguisher should be onboard a vessel with a permanently installed fuel tank

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I think a type B fire extinguisher should be on board a vessel with a permanently installed fuel tank.
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3 years ago

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Gnom [1K]

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3 years ago

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Consider the data presented below. time (s) 0 40 80 120 160 moles of a 0.100 0.067 0.045 0.030 0.020 part a part complete determ

Whitepunk [10]

To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows
time 0 40 80 120 160
moles 0.100 0.067 0.045 0.030 0.020

Q1)
for the first 40 s change of moles ;
= -d[A] / t
= - (0.067-0.100)/40s
= 8.25 x 10⁻⁴ mol/s
for the next 40 s
= -(0.045-0.067)/40
= 5.5 x 10⁻⁴ mol/s
the 40 s after that
= -(0.030-0.045)/40 s
= 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction

rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction

Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation.
Since this is a first order reaction,
b = 1
therefore the reaction is
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
= 8.25 x 10⁻³ s⁻¹

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How many dots should be shown in the Lewis dot diagram of oxygen? (1 point)

deff fn [24]

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