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taurus [48]
3 years ago
6

MARK THE EXTENSIVE PROPERTIES(Check that all apply)

Chemistry
1 answer:
Anarel [89]3 years ago
5 0
I think its just Mass
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In the combustion of hydrogen gas, hydrogen reacts with oxygen from the air to form water vapor.
Yuliya22 [10]
The balanced chemical equation is:

2H2 + O2 ---> 2H2O

We are given the amount of the product produced from the reaction. This will be the starting point for the calculations.

355 g H2O ( 1 mol H2O/ 18.02 g H2O) ( 1 mol O2 / 2 mol H2O ) ( 32 g O2 / 1 mol O2 ) = 315.205 g O2
6 0
3 years ago
Estimate the Calorie content of 65 g of candy from the following measurements. A 15-g sample of the candy is placed in a small a
GrogVix [38]

Answer:

The calorie content of  65g of candy is 326.78 cal

Explanation:

Step 1: Data given

Mass of the candy = 15.00 grams

Mass of the container = 0.325 kg

Mass of water = 1.75kg

0.624 kg at an initial temperature of 15.0°C.

The specific heat of aluminium = 0.22 Cal/kg°C

The specific heat of water = 1 cal/kg°C

Step 2: Calculate calorie content for a 15 gram sample

ΔQ = Σm*c*ΔT

 ⇒ m = mass in grams

⇒ with c= the specific heat in Cal/kg°C

⇒ with ΔT = T2 -T1 = the change in temperatures in °C

ΔQ = m(bomb) * C(aluminium) * ΔT + m(cup) * C(aluminium) * ΔT + m(H2O) * c(H20) * ΔT

ΔQ = (m(bomb) + m(cup)) * c(aluminium)  + m(H2O)*c(H20) ) * ΔT

⇒ with mass of the bomb calorimeter = 0.325 kg

⇒ with mass of the cup = 0.624 kg

⇒ with c(aluminium) = the specific heat of aluminium = 0.22 Cal/kg°C

⇒ with mass of water = 1.75 kg

⇒ with c(water) = the heat capacity of water = 1 Cal/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 53.5 - 15.0 = 38.5 °C

ΔQ = 0.325*0.22*38.5 + 0.624*0.22*38.5 + 1.75*1*38.5

ΔQ = ((0.325 + 0.624)*0.22 + 1.75*1)*38.5

ΔQ = 75.41 cal

Step 3: Calculate the calorie content for a 65 gram sample

For a 65g sample the calorie content will be more or less 4x higher than a 15 gram sample:

ΔQ = 75.41  * (65/15) = 326.78 cal

8 0
2 years ago
A vessel contained N2, Ar, He, and Ne. The total pressure in the vessel was 987 torr. The partial pressures of nitrogen, argon,
xz_007 [3.2K]

Answer:

The partial pressure of neon in the vessel was 239 torr.

Explanation:

In all cases involving gas mixtures, the total gas pressure is related to the partial pressures, that is, the pressures of the individual gaseous components of the mixture. Put simply, the partial pressure of a gas is the pressure it exerts on a mixture of gases.

Dalton's law states that the total pressure of a mixture of gases is equal to the sum of the pressures that each gas would exert if it were alone. Then:

PT= P1 + P2 + P3 + P4…+ Pn

where n is the amount of gases present in the mixture.

In this case:

PT=PN₂ + PAr + PHe + PNe

where:

  • PT= 987 torr
  • PN₂= 44 torr
  • PAr= 486 torr
  • PHe= 218 torr
  • PNe= ?

Replacing:

987 torr= 44 torr + 486 torr + 218 torr + PNe

Solving:

987 torr= 748 torr + PNe

PNe= 987 torr - 748 torr

PNe= 239 torr

<u><em>The partial pressure of neon in the vessel was 239 torr.</em></u>

4 0
3 years ago
How many atoms are in 3 moI of helium
disa [49]
Mark Brainiest please

Answer:
1.8 x 10^24

One mole of helium has 6.02 x 10^23 atoms, thise number is also called Avogadro's number or constant.
If we need number of atoms for 3 moles so simply, multiply 3 by 6.02 x 10^23
so,
Number of atoms for 3 moles = 3 x ()
Number of atoms will be = 1.806x10^24
5 0
3 years ago
Read 2 more answers
How many joules of heat are required to heat 131.0 grams of room temperature water (22 °C) to the boiling point?
vampirchik [111]

100.0g x 4.184J/g/°C x 78°C ΔT = 32,645.2 Joules.

3 0
3 years ago
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