The substance formed on addition of water to an aldehyde or ketone is called a hydrate or a/an:_______
A) vicinal diol
B) geminal diol
C) acetal
D) ketal

Answer:

[tex]\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}[/tex]

Explanation:

1. Density from mass and volume

[tex]\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}[/tex]

2. Volume from density and mass

[tex]V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}[/tex]

3. Mass from density and volume

[tex]\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}[/tex]

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water                = 12.8 mL

Volume of object               = 11.8 mL

[tex]\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}[/tex]

Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

Answer:

Anaphase

Explanation:

The centromere splits during the anaphase of the cell division. Thus, allowing the two linked chromatids to separate.

A typical chromosome is made up of two sister chromatids joined together by a structure known as the centromere. During cell division - at the metaphase stage - the chromosomes align at the equator of the cell, forming the metaphase plate. The spindle from the opposing ends of the cell engages each chromosome at the kinetochore of the centromere.

At the anaphase stage, the centromere splits, leading to the separation of the sister chromatids of each chromosome. The sister chromatids of the same then start migrating in the opposite direction as a result of the shortening of the spindle fiber.

Answer:

The concentration in mol/L =  4.342 mol/L

Explanation:

Given that :

mass of sodium chloride = 25.4 grams

Volume of the volumetric flask = 100 mL

We all know that the molar mass of sodium chloride NaCl = 58.5 g/mol

and number of moles = mass/molar mass

The number of moles of sodium chloride = 25.4 g/58.5 g/mol

The number of moles of sodium chloride = 0.434188 mol

The concentration in mol/L = number of mol/ volume of the solution

The concentration in mol/L = 0.434188 mol/ 100 × 10⁻³ L

The concentration in mol/L =  4.34188 mol/L

The concentration in mol/L =  4.342 mol/L

Answer:

C. Electron cloud

the electron is around 1/2000 times the size of the proton.

If you imagine the proton a a marble in the middle of a football field, the electrons will revolve around the last row

Answer:

0.0002949 moles

Explanation:

Concentration = 2.02×10−2 M

Volume = 14.6 mL = 0.0146 L (Upon converting to litres)

Number of moles = ?

These variables are related by the fllowing equation;

Concentration = Number of moles / Volume

Number of moles = Concentration * Volume

Number of moles = 2.02×10−2 * 0.0146 = 0.0002949 moles

Answer:

There are no forces holding the gas particles together.

Explanation:

Answer:1 )T2=134°C   2) T2=339.48°C. 3)

P=817.59 mmHg.

Explanation:

1.Given ;

pressure, P1 of neon gas = 0.646 atm

temperature, T1 =242oC + 273=515oC

Volume, V1 =515ml

Volume V2= 407ml

temperature , T 2= ?

Solution;

And at constant pressure, the volume cools at V2=407 mL at T2=?

From ideal gas equation, PV=nRT

V/T=constant

therefore

V1/V2=T1/T2 = T2=(V2 xT1)/V1

T2=(407 mL x 515 K)/515 mL= 407K.

T2= 407K -273= 134°C.   recall 0°C=273 K)

2..Given ;

pressure, P1 of neon gas = 0.633 atm

temperature, T1 =261oC + 273=534oC

Volume, V1 =694ml

Volume V2= 796ml

temperature , T 2= ?

Solution;

And at constant pressure, the volume expands  at V2=796mL at T2=?

From ideal gas equation, PV=nRT

V/T=constant

therefore

V1/V2=T1/T2 = T2=(V2 xT1)/V1

T2=(796 mL x 534 K)/694mL= 612.48K.

T2= 612.48K -273= 339.48°C. recall 0°C=273 K

3

Given;

moles of CO2= n=0.962 mol,

temperature T=20°C=20+273 K =293 K,

volume V=21.5 L,

gas constant R at L·mmHg/mol·K= 62.3637 L mmHg mol^-1 K^-1

Using  ideal gas equation PV=nRT

P=nRT/V

P=(0.962 mol)x(62.3637mmHg mol^-1 K^-1)x(293 K)/(21.5L)

P=817.59 mmHg.

Answer:

0.107 mole of SO2.

Explanation:

1 mole of a gas occupy 22.4 L at standard temperature and pressure (STP).

With the above information, we can simply calculate the number of mole of SO2 that will occupy 2.4 L at STP.

This can be obtained as follow:

22.4 L contains 1 mole of SO2.

Therefore, 2.4 L will contain = 2.4/22.4 = 0.107 mole of SO2.

Therefore, 0.107 mole of SO2 is present in 2.4 L at STP.

Answer:

THE MASS OF NITROGEN GAS IN THIS CONDITIONS IS 0.0589 g

Explanation:

In an ideal condition

PV = nRT or PV = MRT/ MM where:

M = mass = unknown

MM =molar mass = 28 g/mol

P = pressure = 2 atm

V = volume = 25 mL = 0.025 L

R = gas constant = 0.082 L atm/mol K

T = temperature = 290 K

n = number of moles

The gas in the question is nitrogen gas

Molar mass of nitrogen gas = 14 * 2 = 28 g/mol

Then equating the variables and solving for M, we have

M = PV MM/ RT

M = 2 * 0.025 * 28 / 0.082 * 290

M = 1.4 / 23.78

M = 0.0589 g

The mass of the nitrogen gas at ideal conditions of 2 atm, 25 mL volume and 290 K temperature is 0.0589 g

Answer:

[tex]\large \boxed{1.503 \times 10^{23}\text{ molecules of Cu(OH)}_{2}}$}[/tex]

Explanation:

You must calculate the moles of Cu(OH)₂, then convert to molecules of Cu(OH)₂.

1. Moles of Cu(OH)₂[tex]\text{Moles of Cu(OH)}_{2} = \text{24.35 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \text{0.2496 mol Cu(OH)}_{2}[/tex]

2. Molecules of Cu(OH)₂[tex]\text{No. of molecules} = \text{0.2496 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= \mathbf{1.503 \times 10^{23}}\textbf{ molecules Cu(OH)}_{2}\\\text{There are $\large \boxed{\mathbf{1.503 \times 10^{23}}\textbf{ molecules of Cu(OH)}_{2}}$}[/tex]

The number of molecules of copper (II) hydroxide in 23.45 g sample has been [tex]\rm \bold {1.45\;\times\;10^2^3}[/tex].

According to the Avogadro number, the number of molecules in a mole of atom has been equivalent to the Avogadro constant. The value of Avogadro constant has been [tex]\rm 6.023\;\times\;10^2^3[/tex].

The moles of a compound has been given as:

[tex]\rm Moles=\dfrac{Mass}{Molar\;mass}[/tex]

The moles in 23.45 g copper (II) hydroxide has been:

[tex]\rm Moles=\dfrac{23.45}{97.562} \\Moles=0.24\;mol[/tex]

The moles of copper (II) hydroxide has been 0.24 mol.

The number of molecules in 0.24 mol sample has been driven by:

[tex]\rm 1\;mol=6.023\;\times\;10^2^3\;molecules\\0.24\;mol=0.24\;\times\;10^2^3\;molecules\\0.24\;mol=1.45\;\times\;10^2^3\;molecules[/tex]

The number of molecules of copper (II) hydroxide in 23.45 g sample has been [tex]\rm \bold {1.45\;\times\;10^2^3}[/tex].

For more information about molecules in a mole of sample, refer to the link:

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Answer:

Following are the answer to this question:

Explanation:

For the reductive amination of its carbonyl group, amino acids could be synthesized by reducing ammunition, which can be synthesized in the given attachment file:

please find the attachment:

Answer:

2.3 × 10⁻⁹

Explanation:

Step 1: Write the reaction for the solution of calcium oxalate

CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)

Step 2: Make an ICE chart

We can relate the molar solubility (S) with the solubility product constant (Ksp) through an ICE chart.

        CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)

I                                0                 0

C                              +S               +S

E                                S                 S

The solubility product constant is:

Ksp = [Ca²⁺] × [C₂O₄²⁻] = S² = (4.8 × 10⁻⁵)² = 2.3 × 10⁻⁹

Answer:

The volume that a mass of 47.2 g would occupy is 0.0334 L

Explanation:

Density is the property that matter, whether solid, liquid or gas, has to compress into a given space. Density is defined as the amount of mass it has per unit volume, that is, the ratio between the mass of a body and the volume it occupies:

[tex]Density=\frac{mass}{volume}[/tex]

This indicates that density is inversely proportional to volume: the smaller the volume occupied by a given mass, the higher the density.

In this case:

Replacing:

[tex]1.413 \frac{g}{mL}=\frac{47.2 g}{volume}[/tex]

Solving:

[tex]volume=\frac{47.2 g}{1.413\frac{g}{mL} }[/tex]

volume=33.40 mL

Being 1,000 mL= 1 L:

volume= 0.0334 L

The volume that a mass of 47.2 g would occupy is 0.0334 L

Answer:

The gas is Methane at 340K

Answer:

CHC12

Explanation:

i am not really sure i am onna do a quick research 4 u tho

Answer:

The mass of ammonium phosphate produced is 14.3g

Explanation:

Full question contains: "What mass of ammonium phosphate is produced by the reaction of 4.9g of ammonia"

Ammonium phosphate ((NH₄)₃PO₄) can be produced by the reaction of phosphoric acid (H₃PO₄) with ammonia (NH₃) as follows:

H₃PO₄ + 3NH₃ → (NH₄)₃PO₄

Where 1 mole of phosphoric acid reacts with 3 moles of ammonia producing 1 mole of ammonium phosphate.

To know how many grams of ammonium phosphate we need to find moles of ammonia that react, and, with the chemical equation we can find moles of ammonium phosphate and its mass as follows:

Moles ammonia (Molar mass: 17.031g/mol):

4.9g × (1mol / 17.031g) = 0.288 moles of ammonia you have in 4.9g

Moles of ammonium phosphate (149.09g/mol) and its mass:

As 0.288 moles of NH₃ are reacting and 3 moles of ammonia produce 1 mole of ammonium phosphate, moles produced are:

Moles (NH₄)₃PO₄:

0.288 moles NH₃ ₓ (1 mol (NH₄)₃PO₄  / 3 mol NH₃) = 0.0959 moles (NH₄)₃PO₄

These moles are, in grams:

0.0959 moles (NH₄)₃PO₄ ₓ (149.09g / mol) = 14.3g ammonium phosphate.

Answer:

Solid sodium hydroxide dissolves in water to form an aqueous solution of ions.

NaOH(s) ⇌ Na+(aq) + OH–(aq) ΔH1 = ?

Solid sodium hydroxide reacts with aqueous hydrochloric acid to form water and an aqueous solution of sodium chloride.

NaOH(s) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH2 = ?

Solutions of aqueous sodium hydroxide and hydrochloric acid react to form water and aqueous sodium chloride.

Na+(aq) + OH–(aq) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH3 = ?

Answer:

A) 3.17 g of Zn

Explanation:

Let's consider the reduction of Zn(II) that occurs in an electrolysis bath.

Zn⁺²(aq) + 2e⁻ → Zn(s)

We can establish the following relations:

The mass of Zn deposited under these conditions is:

[tex]24min \times \frac{60s}{1min} \times \frac{6.5C}{s} \times \frac{1mol\ e^{-} }{96,468C} \times \frac{1molZn}{2mol\ e^{-}} \times \frac{65.38g}{1molZn} = 3.17 g[/tex]

Answer:

A.) 3.17

Explanation:

I got it right in class!

Hope this Helps!! :))

Answer:

Spacecraft’s acceleration (a) = 0.10 N / 500 kg

Explanation:

Given:

Mass of Spacecraft (M) = 500 Kg

Force generate by engine (F) = 0.10 N

Find:

Spacecraft’s acceleration (a)

Computation:

F = Ma

0.10 = 500 (a)

a = 0.10 / 500

Spacecraft’s acceleration (a) = 0.10 N / 500 kg

The expression which can be used to find the spacecraft’s acceleration if its engine created a net force of 0.10 N is  0.10 N/500Kg.

We know that force is the product of the mass a body and its acceleration. The result of motion is the action of an unbalanced force. We have the following information;

Mass of the spacecraft = 500 kg

Force on the engine =  0.10 N

From Newton's law;

F = ma

F = force

m = mass

a = acceleration

a = F/m

acceleration = 0.10 N/500Kg

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Answer:

-170.65

188.8+ 256.8-205.8-(2x205.2)

-170.65 is the entropy change.

Entropy trade is the phenomenon that is the measure of change of disorder or randomness in a thermodynamic gadget. It is associated with the conversion of heat or enthalpy completed in work. A thermodynamic device that has extra randomness means it has high entropy.

Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each extended by using their suitable stoichiometric coefficients, to reap ΔS° for the reaction.

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#SPJ2

Answer:

The equilibrium concentration of ICl is 2.26 M

Explanation:

Chemical equilibrium is a state in which no changes are observed as time passes, despite the fact that the substances present continue to react. This is because chemical equilibrium is established when the forward and reverse reaction take place simultaneously at the same rate.

For the study of chemical equilibrium, the so-called equilibrium constant Kc is useful. Being:

aA + bB ⇔ cC + dD

the equilibrium constant Kc is:

[tex]Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }[/tex]

That is, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

For the reaction:

2 ICl (g) ⇌ I₂ (g) +Cl₂ (g)

the constant Kc is:

[tex]Kc=\frac{[I_{2} ]*[Cl_{2} ]}{[ICl]^{2} }[/tex]

Being Kc =0.110 and the concentration being the amount of moles of solute that appear dissolved in each liter of the mixture and being calculated by dividing the moles of the solute by the liters of the solution:

and replacing in the constant we get:

[tex]0.110=\frac{0.750*0.750}{[ICl]^{2} }[/tex]

Solving, you get the ICl concentration at equilibrium:

[tex][ICl]^{2} =\frac{0.750*0.750}{0.110 }[/tex]

[tex][ICl] =\sqrt{\frac{0.750*0.750}{0.110 }}[/tex]

[ICl]= 2.26 M

The equilibrium concentration of ICl is 2.26 M

Answer:

B) The complex ion is favored over solid silver chloride

C) The free Ag+ ion is unstable.

Explanation:

Hello,

In this case, since the dissociation of solid silver chloride occurs at equilibrium with a neglectable solubility product (very small Ksp), which means that the solid tends to remain undissolved:

[tex]AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)[/tex]

By the addition of ammonia, the following reaction is favored:

[tex]Ag^+(aq)+2NH_3(aq)\rightleftharpoons [Ag(NH_3)_2]^+(aq)[/tex]

Which has a large equilibrium constant, which means that the formation of the complex is assured. In such a way, by addition of more ammonia, more complex will be formed, therefore B) The complex ion is favored over solid silver chloride is true. Moreover, C) The free Ag+ ion is unstable, since they tend to form the complex once they are formed by the solid silver chloride so it readily reacts.

Best regards.

Explanation:

To calculate [H3O+] in the solution we must first find the pH from the [ OH-]

That's

pH + pOH = 14

pH = 14 - pOH

To calculate the pOH we use the formula

pOH = - log [OH-]

And [OH-] = 5.5 × 10^-5 M

So we have

pOH = - log 5.5 × 10^ - 5

pOH = 4.26

Since we've found the pOH we can now find the pH

That's

pH = 14 - 4.26

pH = 9.74

Now we can find the concentration of H3O+ in the solution using the formula

pH = - log H3O+

9.74 = - log H3O+

Find the antilog of both sides

The solution is basic since it's pH lies in the basic region.

Hope this helps you

Answer:

The pH of the solution is 9.06.

Explanation:

The reaction of the dissociation of NH₃ in water is:

NH₃(aq) + H₂O(l)  ⇄  NH₄⁺(aq) + OH⁻(aq)     (1)

[NH₃] - x                     [NH₄⁺] + x     x  

The concentration of NH₃ and NH₄⁺ is:

[tex] [NH_{3}] = \frac{n_{NH_{3}}}{V_{T}} = \frac{C_{i}_{(NH_{3})}*Vi_{NH_{3}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.12 M*0.2 L}{0.2 L + 0.25 L} = 0.053 M [/tex]

[tex] [NH_{4}^{+}] = \frac{C_{i}_{(NH_{4}^{+})*V_{NH_{4}^{+}}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.15 M*0.25 L}{0.2 L + 0.25 L} = 0.083 M [/tex]

From equation (1) we have:

[tex]Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]}[/tex]

[tex] 1.8 \cdot 10^{-5} = \frac{(0.083 + x)*x}{0.053 - x} [/tex]

[tex] 1.8 \cdot 10^{-5}(0.053 - x) - (0.083 + x)*x = 0 [/tex]

By solving the above equation for x we have:

x =  1.15x10⁻⁵ = [OH⁻]

The pH of the solution is:

[tex] pOH = -log([OH^{-}]) = -log(1.15 \cdot 10^{-5}) = 4.94 [/tex]

[tex] pH = 14 - pOH = 14 - 4.94 = 9.06 [/tex]

Therefore, the pH of the solution is 9.06.

I hope it helps you!

Answer:

Explanation:

50 HP = 50 x 746 watt

= 37300 watt

= 37300 J /s

heat produced in one hour = 60 x 60 x 37300 J

= 134280 x 10³ J

latent heat of vaporization = 2260 x 10³ J / kg .

for evaporation of water of mass 1 kg , 2260 x 10³ J of heat is required .

kg of water being evaporated by boiler per hour

= 134280 x 10³ / 2260 x 10³

= 59.41 kg

rate of production of steam

= 59.41 kg / hr .

Answer:

In order from lowest to highest:

Methane < Ethane < Chloroethene < Methanol

i.e: CH4 < CH3CH3 < CH3CH2OH < CH3CH2Cl

Explanation:

Compounds with stronger molecular fore have higher boiling points, thus making the molecules more difficult to pull apart. The presence of chains also increases the molecular dispersion. The dipole force of ethanol makes it have a very high boiling point.

I'm positive this explanation would suffice. Best of luck.

The order of increasing boiling points of the substances listed is; CH4 < CH3CH3 < CH3CH2Cl < CH3CH2OH.

Intermolecular interactions occur between molecules. The boiling point and melting points of substances depends on the nature and magnitude of intermolecular interaction between the molecules of the substance.

The order of increasing boiling points of the substances listed is as follows; CH4 < CH3CH3 < CH3CH2Cl < CH3CH2OH. CH3CH2OH has the highest boiling point due to intermolecular hydrogen bonds in the molecule. Though CH4 and CH3CH3 are both alkanes, CH3CH3 has a higher molecular mass and consequently greater dispersion forces and a higher boiling point.

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Answer:

Answers are in the explanation

Explanation:

It is possible to obtain K of equilibrium of related reactions knowing the laws:

A + B ⇄ C K₁

C ⇄ A + B K = 1 /K₁

The inverse reaction has the inverse K equilibrium

2A + 2B ⇄ 2C K = K₁²

The multiplication of the coefficients of reaction produce a k powered to the number you are multiplying the coefficients

For the reaction:

2 CO2(g) + 4 H2O(g) ⇄ 2 CH3OH(l) + 3 O2(g) K

1) CH3OH(l) + 3/2 O2(g) ⇄ CO2(g) + 2 H2O(g)

This is the inverse reaction but also the coefficients are dividing in the half, that means:

[tex]K_1 = \frac{1}{k^{1/2}} = (1/K)^{1/2}[/tex]

2) CO2(g) + 2 H2O(g) ⇄ CH3OH(l) + 3/2 O2(g)

Here,the only change is the coefficients are the half of the original reaction:

[tex]K_2 = K^{1/2}[/tex]

3) 2CH3OH(l) + 3 O2(g) ⇄ 2 CO2(g) + 4 H2O(g)

This is the inverse reaction. Thus, you have the inverse K of equilibrium:

[tex]K_3 = \frac{1}{K}[/tex]

Answer:

[tex][F^-]_{max}=4x10{-3}\frac{molF^-}{L}[/tex]

Explanation:

Hello,

In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

[tex]Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)[/tex]

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:

[tex][F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}} \\[/tex]

[tex][F^-]_{max}=4x10{-3}\frac{molF^-}{L}[/tex]

Best regards.

Answer:

a.

[tex]m_K=8.056gK\\ \\m_{Cl_2}=4.028gCl_2[/tex]

b.

[tex]m_{Cr}=10.51gCr\\ \\m_{O_2}=4.851gO_2[/tex]

c.

[tex]m_{Sr}=13.88gSr\\\\m_{N_2}=1.479gN_2[/tex]

Explanation:

Hello,

In this case, we proceed via stoichiometry in order to compute the masses of all the reactants as shown below:

a. [tex]2K+Cl_2\rightarrow 2KCl[/tex]

[tex]m_K=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{2molK}{2molKCl}* \frac{39.1gK}{1molK}=8.056gK\\ \\m_{Cl_2}=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{1molCl_2}{2molKCl}* \frac{70.9gCl_2}{1molCl_2}=4.028gCl_2[/tex]

b. [tex]4Cr+ 3O_2\rightarrow 2Cr_2O_3[/tex]

[tex]m_{Cr}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{4molCr}{2molCr_2O_3}* \frac{52gCr}{1molCr_2O_3}=10.51gCr\\ \\m_{O_2}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{3molO_2}{2molCr_2O_3}* \frac{32gO_2}{1molCr_2O_3}=4.851gO_2[/tex]

c. [tex]3Sr(s) + N_2(g) \rightarrow Sr_3N_2[/tex]

[tex]m_{Sr}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{3molSr}{1molSr_3N_2}* \frac{87.62gSr}{1molSr}=13.88gSr\\\\m_{N_2}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{1molN_2}{1molSr_3N_2}* \frac{28gN_2}{1molN_2}=1.479gN_2[/tex]

Regards.