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Finger [1]
3 years ago
13

2) If 100.0 J of heat are added to 20.0 g of water at 30.0°C, what will be the final

Chemistry
1 answer:
anzhelika [568]3 years ago
3 0
The final temperature of the water will be 31.2 °C... i don’t know the second one sorry :(
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As you move to a higher elevation (e.g. in the mountains), the air pressure drops significantly. In turn,
soldi70 [24.7K]
The answer is C ( The evaporation rate of water increases at all temperatures causing an increase in volume.)                           
3 0
3 years ago
A scuba diver that ascends to the surface too quickly can experience decompression sickness, which occurs when nitrogen that dis
Butoxors [25]

Answer:

The factor that will change the volume of the diver's lungs upon reaching the surface is 4

Explanation:

Given data:

Pressure increases 1 atm = 101.325 kPa

34 ft = 10.3632 m

Depth of 102 ft = 31.0896 m

Question: What factor will the volume of the diver's lungs change upon arrival at the surface, V₂/V₁ = ?

The pressure at 31.0896 m:

P_{1} =101.325+(\frac{31.0896}{10.3632} *101.325)=405.3kPa

The factor will the volume of the diver's lungs change upon arrival at the surface:

\frac{V_{2}}{V_{1} } =\frac{P_{1}}{P_{2}} =\frac{405.3}{101.325} =4

5 0
3 years ago
How many grams of ethane gas (C2H6) are in a 12.7 liter sample at 1.6 atmospheres and 24°C? Show all work used to solve this pro
dexar [7]

n = PV/RT

p = 1.6 atm

v = 12.7L

R = 0.0821

T = 24°C which is equivalent to 297.15 degrees k

n = (16 × 12.7) / (0.0821 × 297.15)

n = 20.32 / 24.39

n = 0.83 mol

C = 12.90

H = 1.0079

C2 = 12.010 × 2 = 24.02

H6 = 1.0079 × 6 = 6.0474

C2H6 = 30.0674

Ethane times n which is 30.0674 × 0.83mol

= 24.95 grams of C2H6. Which is Ethane.

4 0
4 years ago
For the reaction, calculate how many grams of the product form when 14.4 g of Br2 completely reacts. Assume that there is more t
lora16 [44]

Answer:

The answer to your question is 21.45 g of KBr

Explanation:

Chemical reaction

                               2K + Br₂   ⇒   2KBr

                                       14.4            ?

Process

1.- Calculate the molecular mass of bromine and potassium bromide

Bromine = 2 x 79.9 = 159.8g

Potassium bromide = 2(79.9 + 39.1) = 238 g

2.- Solve it using proportions

              159.8 g of Bromine ------------ 238 g of potassium bromide                    

                14.4 g of Bromine  ------------  x

                        x = (14.4 x 238) / 159.8

                        x = 3427.2 / 159.8

                        x = 21.45g of KBr

7 0
4 years ago
Five elements named after color
Dmitry [639]
Caesium -bluish(Latin) Chlorine -yellow/green (Greek) Iodine -violet (Greek) <span>Rhodium -rose (Greek) Sulphur - yellow (Arabic)</span>
4 0
3 years ago
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