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2 years ago

CAN SOMEONE HELP ME I WILL GIVE CROWN???!!!!

Chemistry

1 answer:

lutik1710 [3]2 years ago

4 0

Answer:

gravity?

Explanation:

since it's being taken down

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Read the following graduated cylinder. How many ML?

postnew [5]

Answer:

12.8 mL

Explanation:

the bolded bar indicates a count of 5. counting from there would add up to 8/10ths of a mL.

5 0

2 years ago

Water is made of two [Blank] atoms and one [Blank] atom. (please fill in the blanks)

natima [27]

Two [hydrogen] atoms and one [oxygen] atom

5 0

2 years ago

Read 2 more answers

gogolik [260]

Answer: I2 is the Oxidant; while the 2S2O3(-2) is the reductant.

Explanation:

An Oxidant is any substance that oxidizes, or receives electrons from, another; in so doing, it becomes reduced in oxidation number.

A Reductant thus exactly the opposite.

Note that the equation provided shows that Iodine (I2) received an electron to become NEGATIVELY CHARGED:

I2 --> 2I-.

The oxidation number reduced from 0 to -1.

In contrast, the oxidation number of 2S2O3(-2) increases from -4 to -2.

Thus, I2 is the Oxidant; while the 2S2O3(-2) is the reductant.

7 0

3 years ago

If 120.3 mL of water is shaken with oxygen gas at 2.1 atm, it will dissolve 0.0043 g O2. Estimate the Henry's law constant for t

nikklg [1K]

Answer: The Henry's law constant for oxygen gas in water is $1.702\times 10^{-5}g/mL.atm$

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{O_2}=K_H\times p_{O_2}$

where,

$K_H$ = Henry's constant = ?

$C_{O_2}$ = solubility of oxygen gas = $0.0043g/120.3mL$

$p_{O_2$ = partial pressure of oxygen gas = 2.1 atm

Putting values in above equation, we get:

$0.0043g/120.3mL=K_H\times 2.1atm\\\\K_H=\frac{0.0043g}{120.3mL\times 2.1atm}=1.702\times 10^{-5}g/mL.atm$

Hence, the Henry's law constant for oxygen gas in water is $1.702\times 10^{-5}g/mL.atm$

7 0

2 years ago

If 27.3% of a sample of silver-112 decays in 1.52 hours, what is the half-life (in hours to 3 decimal places)?

ICE Princess25 [194]

Answer: The half life of the sample of silver-112 is 3.303 hours.

Explanation:

All radioactive decay processes undergoes first order reaction.

To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:

$k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}$

where,

k = rate constant = ?

t = time taken = 1.52 hrs

$[A_o]$ = Initial concentration of reactant = 100 g

[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g

Putting values in above equation, we get:

$k=\frac{2.303}{1.52hrs}\log \frac{100}{72.7}\\\\k= 0.2098hr^{-1}$

To calculate the half life period of first order reaction, we use the equation:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life period of first order reaction = ?

k = rate constant = $0.2098hr^{-1}$

Putting values in above equation, we get:

$t_{1/2}=\frac{0.693}{0.2098hr^{-1}}\\\\t_{1/2}=3.303hrs$

Hence, the half life of the sample of silver-112 is 3.303 hours.

6 0

3 years ago