According to the law of conservation of mass, what is the same on both sides of a balanced chemical equation?
A. the volume of the substances
B. the subscripts
C. the total mass of atoms
D. the coefficients
Answer:
A balanced equation demonstrates the conservation of mass by having the same number of each type of atom on both sides of the arrow.
Explanation:
Every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. ... Use coefficients of products and reactants to balance the number of atoms of an element on both sides of a chemical equation.
Consider the balanced equation for the combustion of methane.
CH
4
+
2O
2
→
CO
2
+
2H
2
O
All balanced chemical equations must have the same number of each type of atom on both sides of the arrow.
In this equation, we have 1
C
atom, 4
H
atoms, and 4
O
atoms on each side of the arrow.
The number of atoms does not change, so the total mass of all the atoms is the same before and after the reaction. Mass is conserved.
Here is a video that discusses the importance of balancing a chemical equation.
Scientific laws and theories have different jobs to do. A scientific law predicts the results of certain initial conditions. ... In contrast, a theory tries to provide the most logical explanation about why things happen as they do.
Answer:
Explanation:
The answer is 32°F or 0° Celsius.
Hope it helped you.
Answer:
4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.
Explanation:
Half life (t1/2) = 8 days
Original mass (No) = 64 g
Elapsed time (t) = 32 days
Mass remaining (Nt) = ?
Using the half life equation we can obtain the mass remaining (Nt)
Nt = No (1/2) ^t/t1/2
Substituting the values, we have;
Nt = 64 * ( 1/2 ) ^32/8
Nt = 64 * (1/2) ^4
Nt = 64 * 0.0625
Nt = 4 g
So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.
Answer:
45.3°C
Explanation:
Step 1:
Data obtained from the question.
Initial pressure (P1) = 82KPa
Initial temperature (T1) = 26°C
Final pressure (P2) = 87.3KPa.
Final temperature (T2) =.?
Step 2:
Conversion of celsius temperature to Kelvin temperature.
This is illustrated below:
T(K) = T(°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K.
Step 3:
Determination of the new temperature of the gas. This can be obtained as follow:
P1/T1 = P2/T2
82/299 = 87.3/T2
Cross multiply to express in linear form
82 x T2 = 299 x 87.3
Divide both side by 82
T2 = (299 x 87.3) /82
T2 = 318.3K
Step 4:
Conversion of 318.3K to celsius temperature. This is illustrated below:
T(°C) = T(K) – 273
T(K) = 318.3K
T(°C) = 318.3 – 273
T(°C) = 45.3°C.
Therefore, the new temperature of the gas in th tire is 45.3°C
