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hammer [34]
3 years ago
14

Compare the size of I, I+ and I-​

Chemistry
2 answers:
elixir [45]3 years ago
7 0

umm explanation pls so i answee

bogdanovich [222]3 years ago
3 0

Answer:

I+ is bigger

Smartest huh

l=/

Explanation:

I AM SO SMART NO Explanation FOR YOU

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What element has a half-life = to 64
Virty [35]
The best answer is the isotope of strontium which is strontium-85. It has a half-life of about 64 days. The metal strontioum has four stable, naturally occurring isotopes which includes 84Sr (0.56%), 86Sr (9.86%), 87Sr (7.0%) and 88Sr (82.58%).
5 0
3 years ago
Toothpaste is an alkali. How could you use the toothpaste to show that red cabbage is an indicator?
Alecsey [184]

Answer:

PUT THE TOOTHPAST ON CABBAGE AND CHECK IT OUT AFTER A DAY

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3 years ago
An obese man was discovered in his air-conditioned hotel room sitting in a chair in front of the television. The air conditioner
Leno4ka [110]

Answer:

It has been approximately 6 hours after death.

Explanation:

This is because between 2-6 hours after death, the body starts becoming stiff from top to bottom, then spreading to the limbs. Since there is only rigor in his upper body, that would mean that with normal temperature and body conditions, it would be 4 or 5 hours after death. But since he is obese and in cold temperature, there is slower progression of rigor, leading to the maximum time in the first rigor mortis phase, 6 hours.

8 0
3 years ago
Take a cup of water, add sugar, and stir. if the resulting solution contains sugar crystals that do not dissolve, the solution i
Strike441 [17]
When a solvent has as much of the dilute dissolved in it as possible, then it is saturated.

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Hope that helps!
6 0
3 years ago
How much ice (in grams) would have to melt to lower the temperature of 353 mL of water from 26 ∘C to 6 ∘C? (Assume the density o
Rashid [163]

Answer:

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

Explanation:

Heat gain by ice = Heat lost by water

Thus,  

Heat of fusion + m_{ice}\times C_{ice}\times (T_f-T_i)=-m_{water}\times C_{water}\times (T_f-T_i)

Where, negative sign signifies heat loss

Or,  

Heat of fusion + m_{ice}\times C_{ice}\times (T_f-T_i)=m_{water}\times C_{water}\times (T_i-T_f)

Heat of fusion = 334 J/g

Heat of fusion of ice with mass x = 334x J/g

For ice:

Mass = x g

Initial temperature = 0 °C

Final temperature = 6 °C

Specific heat of ice = 1.996 J/g°C

For water:

Volume = 353 mL

Density (\rho)=\frac{Mass(m)}{Volume(V)}

Density of water = 1.0 g/mL

So, mass of water = 353 g

Initial temperature = 26 °C

Final temperature = 6 °C

Specific heat of water = 4.186 J/g°C

So,  

334x+x\times 1.996\times (6-0)=353\times 4.186\times (26-6)

334x+x\times 11.976=29553.16

345.976x = 29553.16

x = 85.4197 kg

Thus,  

<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>

7 0
4 years ago
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