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Colt1911 [192]
3 years ago
6

A balanced chemical reaction obeys the law of...

Chemistry
2 answers:
AveGali [126]3 years ago
7 0
The correct answer is C.Conservation of matter.
Anit [1.1K]3 years ago
4 0

Answer: (C) conservation of matter

Solution: Law of conservation of matter or mass states that' total mass of the reactants should always be equal to the total mass of the product that is the total mass is remained conserved in a chemical reaction.

A balanced chemical equation always follow this law.

For example:

2H_2(g)+O_2(g)\rightarrow 2H_2O(l)

Mass of hydrogen = 1 g/mol

Mass of Oxygen = 16 g/mol

Total mass on the reactants = 2(2×1)+(2×16)= 36g/mol

Total mass on the product side = 2[(2×1) +16] = 36 g/mol

As,

Mass on reactant side = Mass on the product side

Therefore, a balanced chemical reaction follows Law of Conservation of mass.

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If two objects have the same amount of matter, the object with the higher temperature has the higher thermal energy. (science qu
Rom4ik [11]

Answer:

is this a true or false question?

5 0
3 years ago
Read 2 more answers
What is the most likely reason a computer model is used for a black hole?
Rashid [163]

Answer:

Black holes are merely the most exotic example of the general principle that ... quantum computer stores bits on protons and uses magnetic fields to flip them. ... Powered by Standard Model software, the universe computes

Explanation:because they are mostly exotic

6 0
3 years ago
A hot air balloon is filled with 15 moles helium gas and 5 moles nitrogen gas. What is the volume of the balloon at 1.01 atm and
kvasek [131]

<u>Given:</u>

Moles of He = 15

Moles of N2 = 5

Pressure (P) = 1.01 atm

Temperature (T) = 300 K

<u>To determine:</u>

The volume (V) of the balloon

<u>Explanation:</u>

From the ideal gas law:

PV = nRT

where P = pressure of the gas

V = volume

n = number of moles of the gas

T = temperature

R = gas constant = 0.0821 L-atm/mol-K

In this case we have:-

n(total) = 15 + 5 = 20 moles

P = 1.01 atm and T = 300K

V = nRT/P = 20 moles * 0.0821 L-atm/mol-K * 300 K/1.01 atm = 487.7 L

Ans: Volume of the balloon is around 488 L


3 0
3 years ago
A very old tree limb contains an amount of carbon-14 that is approximately 1/8 of the current atmospheric 14C levels.. Calculate
Tomtit [17]

A.       The radioactive decay equation is N = N0e^{-ln(2)*t/T }

where T is the half-life (5730 years), N0 is the number of atoms at time t = 0 and N is the number at time t.

Rewriting this as:

(N/N0) = e^{-ln(2)*t/T }

Since N = (1/8) N0 and substituting known values:

1/8 = e^{-ln(2)*t/5730}

Taking ln of both sides:

ln(1/8)= -ln(2)*t/5730

t = - 5730 * ln(1/8) / ln (2)

t = 17,190 years

The tree was cut down 17,190 years ago.

B.   N0 = 1,500,000 carbon-14 atoms

Since N = (1/8) N0

N = 187,500 carbon atoms left

3 0
3 years ago
I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0
3 years ago
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