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olya-2409 [2.1K]
2 years ago
6

Brock wants to see a full cycle of one high tide to the next. About how long will Brock have to wait?

Chemistry
1 answer:
german2 years ago
8 0

Answer:

5 minutes because he is good guy

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What di you think you would see happen in the bucket as ice comes into contact with high kinetic energy molecules?
cricket20 [7]

Answer:The ice will melt

Explanation:

5 0
3 years ago
Please help me
denis23 [38]

Answer:

it would be 335

Explanation:

3 0
3 years ago
The Henry's law constant (kH) for O2 in water at 20°C is 1.28e-3 mol/l atm. How many grams of O2 will dissolve in 3.5 L of H2O t
Sophie [7]

Answer : The mass of O_2 dissolved will be, 0.2365 grams

Explanation :

First we have to calculate the concentration of O_2.

As we know that,

C_{O_2}=k_H\times p_{O_2}

where,

C_{O_2} = concentration of O_2 = ?

p_{O_2} = partial pressure of O_2 = 1.65 atm

k_H = Henry's law constant = 1.28\times 10^{-3}mole/L.atm

Now put all the given values in the above formula, we get:

C_{O_2}=(1.28\times 10^{-3}mole/L.atm)\times (1.65atm)

C_{O_2}=2.112\times 10^{-3}mole/L

The concentration of O_2 = 2.112\times 10^{-3}mole/L

Now we have to calculate the moles of O_2

\text{Moles of }O_2=\text{Concentration of }O_2\times \text{volume of solution}

\text{Moles of }O_2=(2.112\times 10^{-3}mole/L)\times (3.5L)=7.392\times 10^{-3}mole

Now we have to calculate the mass of O_2

\text{Mass of }O_2=\text{Moles of }O_2\times \text{Molar mass of }O_2

\text{Mass of }O_2=(7.392\times 10^{-3}mole)\times (32g/mole)=0.2365g

Therefore, the mass of O_2 dissolved will be, 0.2365 grams

4 0
3 years ago
A compound sample contains 60.87% c, 4.38% h, and 34.75% o by mass. it has a molar mass of 276.2 g/mol. what are the empirical a
Allushta [10]

Step 1) Assume we have 100 grams of the compound. Thus, we're starting with 60.87 grams C, 4.38 grams H, and 34.75 grams O.

Step 2) Convert the masses of each to moles.

60.87 grams C=5.068276436 mol C

4.38 grams H=4.345238095 mol H

34.75 grams O=2.171875 mol O

step 3) Determine your simplest whole-number ratio of moles by dividing each number of moles by the smallest number of moles. (In this case, the smallest number of moles is 2.171875 mol O)

O: \frac{2.171875}{2.171875} = 1

H:\frac{4.345238095}{2.171875} = 2

C: \frac{5.068276436}{2.171875} = 2.33

Step 4) Whenever you're doing an empirical formula problem, and you run into a number that ends with .33, rather than rounding down to a whole number, you need to multiply all your ratios by 3 to obtain whole numbers. Thus, you will get:

O=3

H=6

C=7

Step 5) write the empirical formula using the ratios.

The empirical formula is: C₇H₆O₃

Step 6) The subscripts in the molecular formula of a substance are always whole-number multiples of the subscripts in its empirical formula. This multiple is found by dividing the molecular weight (which was given to us in the problem: 276.2 amu) by the empirical formula weight (C₇H₆O₃ =138.118 amu).

\frac{276.2}{138.118} = 2

Step 7) simply multiply the subscripts in the empirical formula by the multiple, 2, and you will get the molecular formula.

The molecular formula is: C₁₄H₁₂O₆

6 0
3 years ago
Please help to me solve these ! :)
emmainna [20.7K]
I. The solubility of NaCl at 25 degrees C would be between the solubilities at 20 and 30 degrees C. A reasonable answer would be 36 grams/100 g water
ii. From the table, it’s clear that the salts are more soluble at higher temperatures, indicating that an increase in temperature increases solubility.
iii. At 50 degrees C, a saturated ammonium chloride solution will have 50.6 grams of salt per 100 g water. At 20 degrees C, the solution can hold only 37.3 grams of salt per 100 g water. Thus, 13.3 grams of salt will precipitate per 100 grams of water.
4 0
4 years ago
Read 2 more answers
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