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tester [92]
2 years ago
13

A beaker with water and the surrounding air are all at 24°C. After ice cubes are placed in the water, heat is transferred from:

Chemistry
2 answers:
Katen [24]2 years ago
7 0
The answer is (3) the water to ice cubes.

As the ice cubes should be at a temperature of about 0 degree ( freezing point) , at the same time the temperature of water is 24 degree. Thus, heat is transferred from water to ice cubes.
xxTIMURxx [149]2 years ago
5 0
(3) the water to the ice cubes
  This is so because the ice cubes are colder compared to their surrounding objects.
They absorb the heat from the water, cooling it, therefore vanishing(melting).
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How many cm 3 are in 0.14 m 3?
nlexa [21]

Answer:

There are 14 cm in 0.14 m

5 0
3 years ago
At a certain temperature, the solubility of n2 gas in water at 3.08 atm is 72.5 mg of n2 gas/100 g water . calculate the solubil
8090 [49]

According to Henry's law, solubility of solution is directly proportional to partial pressure thus,

\frac{S_{1}}{P_{1}}=\frac{S_{2}}{P_{2}}

Solubility at pressure 3.08 atm is 72.5/100, solubility at pressure 8 atm should be calculated.

Putting the values in equation:

\frac{0.725}{3.08}=\frac{S_{2}}{8}

On rearranging,

S_{2}=\frac{0.725\times 8}{3.08}=1.88

Therefore, solubility will be 1.88 mg of N_{2} gas in 1 g of water or, 188 mg of tex]N_{2}[/tex] gas in 100 g of water.

3 0
2 years ago
A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

8 0
2 years ago
Please awnser the fill in blanks don’t mind the top or bottom pls help !!!!
zysi [14]
Can’t see the problems
7 0
2 years ago
Which material is very strong and tough but shows very little elongation as it absorbs energy?
natita [175]
Kevlar would be the answer!! enjoy the rest of your day broskies!!!
7 0
2 years ago
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