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Reil [10]
3 years ago
11

I have 0.92 moles of water (H20). What is the mass of the water?​

Chemistry
1 answer:
strojnjashka [21]3 years ago
5 0

Ans:

16.57 g

Explanation:

If you have 0.92 moles of water you would take your moles and multiply it by its molar mass (18.01528 g/mol) to find your specific mass.

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The specific heat of aluminum is 0.897j/(g•°C). If a 22.6 g sample of aluminum is heated from 25°C to 250°C, then how much heat
Oxana [17]

Q=mcat

=(22.6)(.897)(250-25)

4,561.245 or 4.5 * 10^-3

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3 years ago
You have a particle of length 68 nm. What is this in meters? ( need help asap)
Bess [88]

Answer:

B

Explanation:

4 0
2 years ago
In water, hydrogen bonds are best described as _____.
Elina [12.6K]

Answer: Option D) covalent bonds between water molecules

In water, hydrogen bonds are best described as covalent bonds between water molecules

Explanation:

The hydrogen bonds between water molecules are covalent bonds because they are formed when oxygen attract the lone electron in hydrogen, thus resulting in the formation of a partially negative charge on the oxygen atom and a partially positive charge on two hydrogen atoms

Thus, the sharing of electrons between oxygen and hydrogen atoms is responsible for the covalent bonds between water molecules

7 0
2 years ago
Please help im really stressed and doing 2 tests at the same time
DanielleElmas [232]
The answer is A, good luck
7 0
3 years ago
Read 2 more answers
A 110.0-mL sample of a solution that is 3.0×10−3 M in AgNO3 is mixed with a 230.0- mL sample of a solution that is 0.10 M in NaC
almond37 [142]

Answer:

[Ag⁺] = 0.0666M

Explanation:

For the addition of Ag⁺ and CN⁻, the (Ag(CN)₂⁻ is produced, thus:

Ag⁺ + 2CN⁻  ⇄  Ag(CN)₂⁻

Kf = 1x10²¹ = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺]

As initial concentrations of Ag⁺ and CN⁻ are:

[Ag⁺] = 0.110L × (3.0x10⁻³mol / L) = 3.3x10⁻⁴mol / (0.110L + 0.230L) = 9.7x10⁻⁴M

[CN⁻] = 0.230L × (0.1mol / L) = 0.023mol / (0.110L + 0.230L) = 0.0676M

The equilibrium concentrations of each compound are:

[CN⁻] = 9.7x10⁻⁴M - x

[Ag⁺] = 0.0676M - x

[Ag(CN)₂⁻] = x

<em>Where x is reaction coordinate</em>

Replacing in Kf formula:

1x10²¹ = [x] / [9.7x10⁻⁴M - x]² [0.0676M - x]

1x10²¹ = [x] / 6.36048×10⁻⁸ - 0.000132085 x + 0.06954 x² - x³

-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = x

-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = 0

Solving for x:

X = 9.614x10⁻⁴M

Thus, equilibrium concentration of Ag⁺ is:

[Ag⁺] = 0.0676M - 9.614x10⁻⁴M = <em>0.0666M</em>

6 0
2 years ago
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