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weeeeeb [17]
3 years ago
5

A 500.0 g sample of aluminium, I initially at 25.0 degrees, absorbs heat from its surroundings and reaches a final temperature o

f 90.7 degrees. How much heat (in KJ) has been absorbed by the sample? To one decimal place
Specific heat= 0.9930j g-1 K-1 for aluminium
Chemistry
1 answer:
babunello [35]3 years ago
3 0
Q= (500.0)(0.9930)(90.7 - 25.0)
Q= 32620.05
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If a pharmacist adds 10 ml of purified water to 30 ml of a solution having a specific gravity of 1.30, calculate the specific gr
xxMikexx [17]

The specific gravity of a sample is the ratio of the density of the sample with respect to one standard sample. The standard sample used in specific gravity calculation is water whose density is 1 g/mL. The solution having specific gravity 1.30 is the density of the sample that is 1.30 g/mL. Thus the weight of the 30 mL sample is (30×1.30) = 39 g.

Now the mass of the 10 mL of water is 10 g as density of water is 10 g/mL. Thus after addition the total mass of the solution is (39 + 10) = 49g and the volume is (30 + 10) = 40 mL. Thus the density of the mixture will be \frac{49}{40}=1.225 g/mL. Thus the specific gravity of the mixed sample will be 1.225 g/mL.

6 0
3 years ago
How do you balance redox equations in acidic solutions?
Anuta_ua [19.1K]

Answer:

First, balance the half-reactions

Second, equalize the electrons

Third,add two reaction equations to get final answer

Explanation:

For example

H₂C₂0₄ + MnO⁻₄ ---------->CO₂+Mn²⁺

(i) Balancing the half reactions

H₂C₂O₄-------->2CO₂+2H⁺+2e⁻

5e⁻ +8H⁺+MnO₄⁻----------->Mn²⁺+4H₂O

(ii)

Equalizing the electrons

5H₂C₂O₄--------->10CO₂+10H⁺+10e⁻  ---here there is a factor of 5

10e⁻+16H⁺+2MnO₄⁻--------->2Mn²⁺+8H₂O -----here there is a factor of 2

(iii)

Add the two where electrons and some Hydrogen ions will cancel out

5H₂C₂O₄+6H⁺+2MnO₄⁻---->10CO₂+2Mn²⁺+8H₂O

7 0
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Andru [333]
The answer should be false. Elements contain only one atom.
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3 years ago
A 0.89% (w/v) sodium chloride solution is referred to as physiological saline solution because it has the same concentration of
maks197457 [2]
1) 0.89% m/v = 0.89 grams of NaCl / 100 ml of solution

=> 8.9 grams of NaCl in 1000 ml of solution = 8.9 grams of NaCl in 1 liter of solution

2) Molarity = M = number of moles of solute / liters of solution

=> calculate the number of moles of 8.9 grams of NaCl

3) molar mass of NaCl = 23.0 g /mol + 35.5 g/mol = 58.5 g / mol

4) number of moles of NaCl = mass / molar mass = 8.9 g / 58.5 g / mol = 0.152 mol

5) M = 0.152 mol NaCl / 1 liter solution = 0.152 M

Answer: 0.152 M
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