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e-lub [12.9K]
3 years ago
12

How many moles are present in 2.95 x1024 molecules of water (H20)

Chemistry
1 answer:
DerKrebs [107]3 years ago
4 0

ANSWER: 3020.8 moleclues of water (h20)           Explanation I did the math and the quiz

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Name the SI base units that are important in chemistry, and give the SI units for expressing the following: (a)length, (b) volum
kirza4 [7]

Answer and Explanation:

The basic unit which are that are important in chemistry are meter, kilogram ,mol,m^3

Candela which is the unit of luminous of intensity is not so important in physics

(a) SI unit of length is meter (m)

(b) Si unit of volume is m^3

(c) Si unit of mass is kilogram (kg)

(d) SI unit of time is second (s)

(e) SI unit of temperature is kelvin (K)

4 0
3 years ago
The cube has a mass of 72.9 g.<br><br> What is its density in g/cm3? <br><br> What substance is it?
sashaice [31]
If the cube is 3 cm on each side, then it has a volume of 27 cm^3 (3 x 3 x 3). Density is mass divided by volume, so its density is 72.9/27 = 2.7 g/cm^3. 

<span>Going by the density, the cube is made of Aluminium - density is a fairly unique quantity</span>
6 0
3 years ago
What is the ph of a solution labeled .30 trimethylamine k for trimethylamine is 7.42 x 10^4
valentina_108 [34]

Answer:the pH is 12

Explanation:

First We need to understand the structure of trimethylamine

Due to the grades of the bond in the nitrogen with a hybridization sp3 is 108° approximately, then is generated a dipole magnetic at the upper side of the nitrogen, this dipole magnetic going to attract a hydrogen molecule of the water making the water more alkaline

C3H9N+ H2O --> C3H9NH + OH-

k=\frac{[C3H9NH]*[OH-]}{[C3H9N]}

Then:

The concentration of the trimethylamine is 0.3 and the concentration of the ion C3H9NH is equal to the OH- relying on the stoichiometric equation. We could find the concentration of the OH- ion with the square root of the multiplication between k and the concentration of trimethylamine

[OH-]=\sqrt{ 0.3*7.42x10^{-4}}

[OH-]=0.01

pH=14-(-log[OH-])

pH=12

5 0
3 years ago
A sample of oxygen gas is compressed from 30.6 L to 1.8 L at constant temperature pressure of 1.8 atm. Calculate the amount of e
ad-work [718]

Answer:

the change in the internal energy of the system is 3,752.67 J

Explanation:

Given;

initial volume of the gas, V₁ = 30.6 L

final volume of the gas, V₂ = 1.8 L

constant pressure of the gas, P = 1.8 atm

Energy released by the system, Q = 1.5 kJ = 1,500 J

Apply pressure-volume work equation, to determine the work done on the gas;

w = -PΔV

w = -P(V₂ - V₁)

w = - 1.8 atm(1.8 L - 30.6 L)

w = 51.84 L.atm

w = 51.84 L.atm x 101.325 J/L.atm

w = 5,252.67 J

The change in the internal energy of the system is calculated as;

ΔU = Q + w

Since the heat is given out, Q = - 1,500 J

ΔU = -1,500 J  +  5,252.67 J

ΔU = 3,752.67 J

Therefore, the change in the internal energy of the system is 3,752.67 J

3 0
3 years ago
how many molecule of carbon dioxide are needed to react with excess iron oxide to produce 11.6 g of iron
lesya692 [45]

Answer:

0.16 moles of Carbon

Explanation:

The balanced reaction equation:

2Fe_{2}O_{3} + 3C → 4Fe + 3CO_{2}↑

The mole ratio of Carbon to Iron is 3 : 4 (since Fe2O3 is in excess)

i.e 3 moles of C produces 4 moles of Fe.

If 1 mole of Fe - 55.8g of Fe

? moles - 11.6g of Fe

= \frac{11.6}{55.8} = 0.208 moles

But 3 moles of C - 4 moles of Fe

? moles of C - 0.208 moles of Fe

= \frac{3 *0.208}{4} = 0.16 moles of carbon.

I hope this explanation was clear and useful.

5 0
3 years ago
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