Answer:
The chemical equation needs to be balanced so that it follows the law of conservation of mass.
Explanation:
Yes, the law of conservation of mass holds.
Explanation:
In every chemical reaction, mass is always conserved. This implies that in chemical reaction, the process proceeds maintaining the same set of atom in the same proportion without new ones forming.
- From the description given, decomposition of the Silver carbonate will produce a silver residue with the given mass.
- The other mass that seems lost can be examined to be given off as carbon dioxide gas which is the other product of the reaction and oxygen.
- Therefore, since the products are silver, carbon dioxide and oxygen, the remaining mass is that of the carbon dioxide and oxygen. It is not lost.
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Answer:
![\boxed {\tt mass=29.25 \ grams}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Ctt%20mass%3D29.25%20%5C%20grams%7D)
Explanation:
The density formula is mass over volume.
![d=\frac{m}{v}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7Bm%7D%7Bv%7D)
Rearrange the formula for the mass, m. Multiply both sides of the formula by v.
![d*v=\frac{m}{v}*v](https://tex.z-dn.net/?f=d%2Av%3D%5Cfrac%7Bm%7D%7Bv%7D%2Av)
![d*v=m](https://tex.z-dn.net/?f=d%2Av%3Dm)
Mass can be found by multiplying the density and volume. The density is 0.65 grams per milliliter and the volume is 45.0 cubic centimeters.
- A cubic centimeter is equal to a milliliter.
- Therefore, 45 cubic centimeters also equals 45 milliliters.
![d= 0.65 \ g/mL\\v= 45 \ mL](https://tex.z-dn.net/?f=d%3D%200.65%20%5C%20g%2FmL%5C%5Cv%3D%2045%20%5C%20mL)
Substitute the values into the formula.
![0.65 \ g/mL * 45 \ mL=m](https://tex.z-dn.net/?f=0.65%20%5C%20g%2FmL%20%2A%2045%20%5C%20mL%3Dm)
Multiply. Note the milliliters, or mL will cancel out.
![0.65 \ g * 45=m](https://tex.z-dn.net/?f=0.65%20%5C%20g%20%2A%2045%3Dm)
![29.25 \ g=m](https://tex.z-dn.net/?f=29.25%20%5C%20g%3Dm)
The mass of the wood is 29.25 grams.
Answer:
7.35atm
Explanation:
Data obtained from the question include:
V1 = 28L
T1 = 42°C = 42 + 273 = 315K
P1 =?
V2 = 49L
T2 = 27°C = 27 + 273 = 300K
P2 = 4atm
Using P1V1/T1 = P2V2/T2, the original pressure can be obtained as follows:
P1V1/T1 = P2V2/T2
P1 x 28/315 = 4 x 49/300
Cross multiply to express in linear form
P1 x 28 x 300 = 315 x 4 x 49
Divide both side by 28 x 300
P1 = (315 x 4 x 49) /(28 x 300)
P1 = 7.35atm
Therefore, the original pressure is 7.35atm