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3 years ago

The law of conservation of mass is applicable to

Chemistry

1 answer:

katrin2010 [14]3 years ago

3 0

Answer:

The law of conservation of mass states that in a closed system, mass is neither created nor destroyed during a chemical or physical reaction. The law of conservation of mass is applied whenever you balance a chemical equation.

Explanation:

According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

The law of conservation of mass is useful for a number of calculations and can be used to solve for unknown masses, such the amount of gas consumed or produced during a reaction.

It is applicable in a chemical when the the mass of the products in a chemical reaction is equal to the mass of the reactants.

But it is not applicable in a nuclear fusion as some of the mass is generated as energy.

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Isoflavones, quercetin and anthocyanins are all what type of phytochemicals

kykrilka [37]

Isoflavones, quercetin, and anthocyanin are all Flavonoids.

8 0

3 years ago

Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa

drek231 [11]

Answer:

1. the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c ( $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$ )

Explanation:

To find -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

So,

Before filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 117 kPa

⇒Absolute pressure ( p1 ) = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 212 kPa

⇒Absolute pressure (p2) = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. $V_{1}$ = $V_{2}$

As we know that, P ∝ M

⇒ $\frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378$ ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒ $\frac{P}{T}$ ∝ M

⇒ $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$

So, The correct relation is c option.

6 0

3 years ago

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.8

Lubov Fominskaja [6]

Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

Answer:

The pK_a value is $pK_a =7.82$

Explanation:

From the question we are told

The volume of base is $V_B = 26.mL = 0.0260L$

The pH of solution is $pH = 7.82$

The concentration of the acid is $C_A = 0.1M$

From the pH we can see that the titration is between a strong base and a weak acid

Let assume that the the volume of acid is $V_A = 18mL= 0.018L$

Generally the concentration of base

$C_B = \frac{C_AV_A}{C_B}$

Substituting value

$C_B = \frac{0.1 * 0.01800}{0.0260}$

$C_B= 0.0692M$

When 13mL of the base is added a buffer is formed

The chemical equation of the reaction is

$HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}$

Now before the reaction the number of mole of base is

$No \ of \ moles[N_B] = C_B * V_B$

Substituting value

$N_B = 0.01300 * 0.0692$

$= 0.0009 \ moles$

Now before the reaction the number of mole of acid is

$No \ of \ moles = C_B * V_B$

Substituting value

$N_A = 0.01800 *0.1$

$= 0.001800 \ moles$

Now after the reaction the number of moles of base is zero i.e has been used up

this mathematically represented as

$N_B ' = N_B - N_B = 0$

The number of moles of acid is

$N_A ' = N_A - N_B$

$= 0.0009\ moles$

The pH of this reaction can be mathematically represented as

$pH = pK_a + log \frac{[base]}{[acid]}$

Substituting values

$7.82 = pK_a +log \frac{0.0009}{0.0009}$

$pK_a =7.82$

8 0

3 years ago

Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ∆H° for the r

Kazeer [188]

Answer:

∆H° rxn = - 93 kJ

Explanation:

Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.

We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.

N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)

1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)

∆H° rxn = ∑ H bonds broken - ∑ H bonds formed

∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]

∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ

be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .

8 0

3 years ago

Most favourable conditions for electrovalency are

anygoal [31]

Answer:

I think

Explanation:

low charge of ions, large cation and small anion

7 0

3 years ago