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1 year ago

11. Which is another way of saying the law of conservation of mass?

Chemistry

1 answer:

Vlada [557]1 year ago

6 0

Answer:

all of the above

Explanation:

The Law of Conservation of Mass states that matter is neither created nor destroyed. It also states that in a chemical reaction, the total mass of the reactants equals the total mass of the products.

The Law of Conservation of Matter states that matter can change form through physical and chemical changes but matter is conserved. The same amount exists before and after the change. Matter is not created or destroyed.

So the answer will be all of the above.

Hope that helps.

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The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

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

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

2 years ago

Motohiro felt chilled by the rainy afternoon. He decided to make a cup of hot tea. He let the tea leaves steep too long and the

kati45 [8]

Answer: the correct one would be option B.

Explanation:

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2 years ago

Convert 1.21 kg to grams

MissTica

Answer:

1210 grams

Explanation:

Message me for extra info.

snap- parkguy786

7 0

2 years ago

Read 2 more answers

doctor has ordered that a patient be given 20 g of glucose, which is available in a concentration of 70.00 g glucose/1000.0 mL o

butalik [34]

The patient should be given 285.71 ml.
1000 ml contains 70 gr glucose.
x contains 20 gr glucose.

x=1000*20/70

8 0

2 years ago

Calculate the molality of a 20.0% by mass ammonium sulfate (nh4)2so4 solution. the density of the solution is 1.117 g/ml.

olasank [31]

Hello!

We have the following data:

m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)

First we find the solute mass (m1), knowing that:

20% m/m = 20g/100mL

20 ------ 100 mL (0,1 L)
y g --------------- 1 L

y = 20/0,1
y = 200 g --> m1 = 200 g

Let's find Solute's Molar Mass, let's see:

M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

We must find the volume of the solvent and therefore its mass (m2), let us see:

d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?

$d = \dfrac{m}{V} \to V = \dfrac{m}{d}$

$V = \dfrac{200\:\diagup\!\!\!\!g}{1,117\:\diagup\!\!\!\!g/mL} \to V = 179\:mL\:(volumen\:of\:solute)$

The solvent volume will be:

1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

Then the mass of the solvent is:

m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{m_1}{M_1*m_2}$

$\omega = \dfrac{200}{132*0,821}$

$\omega = \dfrac{200}{108,372}$

$\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

_________________________________
_________________________________

Another way to find the answer:

We have the following data:

W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

Let's find the number of mols (n), let's see:

 $n = \dfrac{m_1}{M_1}$

$n = \dfrac{200}{132}$

$n \approx 1,5\:mol$

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{n}{m_2}$

$\omega = \dfrac{1,5}{0,8}$

$
\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

I hope this helps. =)

7 0

2 years ago