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ddd [48]
3 years ago
5

Please answer these about Charles law

Chemistry
1 answer:
NNADVOKAT [17]3 years ago
6 0

Answer:

1. V2.

2. 299K.

3. 451K

4. 0.25 x 451 = V2 x 299

Explanation:

1. The data obtained from the question include:

Initial volume (V1) = 0.25mL

Initial temperature (T1) = 26°C

Final temperature (T2) = 178°C

Final volume (V2) =.?

2. Conversion from celsius to Kelvin temperature.

T(K) = T (°C) + 273

Initial temperature (T1) = 26°C

Initial temperature (T1) = 26°C + 273 = 299K

3. Conversion from celsius to Kelvin temperature.

T(K) = T (°C) + 273

Final temperature (T2) = 178°C

Final temperature (T1) = 178°C + 273 = 451K

4. Initial volume (V1) = 0.25mL

Initial temperature (T1) = 299K

Final temperature (T2) = 451K

Final volume (V2) =.?

V1 x T2 = V2 x T1

0.25 x 451 = V2 x 299

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Liono4ka [1.6K]

Answer:

d and e - Sodium and antimony

Explanation:

The atomic numbers remain the same, while the mass numbers change (because neutrons are being added or taken away).

sodium has an atomic number of 19 and a mass number of 39 - in d, it has an atomic number of 19 but a mass number of 40. therefore, it is an isotope

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6 0
3 years ago
At what the pressure, in atmospheres, that O2, gas at 75°C has a density of 2.00 g/L​
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4 years ago
What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP?
BigorU [14]
Mg reaction with O₂ gas will produce MgO so the equation will be
2Mg+O₂⇒2MgO. (You have to find the equation in order two figure out the number of moles of O₂ that will react with 1 mole of MgO).

The first step is to find the number of moles of Mg in 4.03g of Mg.  You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg.  Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg.  To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂.  From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP.  Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.

I hope this helps. Let me know in the comments if anything is unclear.
6 0
4 years ago
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Heating galactose, a monosaccharide sugar, in the presence of excess oxygen produces carbon dioxide gas and water vapor. Classif
artcher [175]

Answer:

it's a combustion reaction

6 0
3 years ago
Consider the reaction between acetylene, c2h2, and oxygen in a welding torch: 2c2h2(g) + 5o2(g) → 4co2(g) + 2h2o(g) if 5.4 moles
fredd [130]
<span>                                      2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
   from the reaction     2 mol                              4 mol
from the problem      5.4 mol                         10.8 mol

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Answer is  C. 4.8*10² g.
4 0
3 years ago
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