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salantis [7]
2 years ago
6

If 10.0 g water at 0.0°C is mixed with 20.0 g of water at 30.0°C what is the final temperature of the mixture?

Chemistry
1 answer:
Bond [772]2 years ago
7 0

Answer:

1.67 gradius Celcius

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

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How much energy does it take to melt 2 kg of ice? (Refer to table of<br> constants for water.)
ioda
It would be C

2 kg x 1000 g/kg x 1mol/18.02 x 6.03 kj/mol = 669kj


5 0
2 years ago
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Which statement accurately describes dark matter?
Mandarinka [93]

Answer:

C. It does not emit electromagnetic radiation.

Explanation:

Right now, Dark Matter is only a theory. Scientist proposed this to counter some of the strange phenomenon with matter in space.

Scientists know little about dark matter. Some say it's one of the driving forces of the universe. Currently, scientists have no way of measuring or identifying dark matter.

5 0
2 years ago
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What mass of propane could burn in 48.0 g of oxygen? C3H8 + 5O2 → 3CO2 + 4H2O
Ipatiy [6.2K]

Answer:

Mass = 13.23 g  

Explanation:

Given data:

Mass of oxygen = 48.0 g

Mass of propane burn = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂     →      3CO₂ + 4H₂O

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 48.0 g/ 32 g/mol

Number of moles = 1.5 mol

now we will compare the moles of propane and oxygen.

              O₂           :          C₃H₈

               5            :            1

             1.5            :          1/5×1.5 = 0.3 mol

Mass of propane burn:

Mass = number of moles × molar mass

Mass = 0.3 mol × 44.1 g/mol

Mass = 13.23 g  

6 0
2 years ago
What is transferred by all types of waves? A. electrons B. energy C. heat D. matter
saveliy_v [14]
<span>B. energy
hope it helps
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6 0
3 years ago
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58. A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains 5.0% CO2, 12.0% O2,
erica [24]

<em><u>Answer and Explanation:</u></em>

Greetings!

Let's~answer~your~question!

Partial ~pressure ~of ~gas ~can ~be ~directly~ calculated ~by ~multiplying ~the~ percentage\\ of~ pressure~ of~ gases~ to ~the ~total ~pressure.

\boxed{Pgas~ = ~P~total~ * \% ~P ~of ~gas}

<em><u>For % of N2 gas: </u></em>

<em><u /></em>100\% - (5\% + 12\%) = 83\% ~N2<em><u /></em>

<em><u /></em>

<em><u /></em>PN_2 ~= ~146~ atm~ *~ 0.83 ~ = 121.18 ~atm<em><u /></em>

<em><u /></em>PO_2 ~= ~146~ atm~ *~ 0.12~ = 17.52 ~atm<em><u /></em>

<em><u /></em>~PCO_2~ = ~146~ atm ~* 0.05 = 7.3 ~atm<em><u /></em>

3 0
3 years ago
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