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maks197457 [2]
3 years ago
7

How many atoms are there in 2.3 grams of zirconiumn?

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

Answer:

1.51834×10^22

Explanation:

Number of atom=Mass/Molar Mass × avogadros constant

2.3g/91.224g/mol × 6.203×10^23

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What is the molecular formula of a compound with the empirical formula CH and a molar mass of 39.01 grams per mole? a.CH b. C2H2
likoan [24]
So knowing the mass of C and H, C=12g and H=1g. We can have 3 C's (3*12=36) and 3 H's (3*1=3) so 36+3 is 39. Plus C3H3 has an empirical formula of CH since C3H3 can be simplified 
3 0
4 years ago
Calculate the mass in grams for each of the following.1.13×1024 atoms of argon, Ar.
natta225 [31]
250g
Explanation: 6.022 times 1023 argon atoms have a mass of 39.95
3 0
4 years ago
4. How many milligrams are in 5.25 x 10-13 kg?<br><br> the “-13” is an exponent
rusak2 [61]

5. 25 x 10⁻⁷mg

Explanation:

This is mass conversion from mg to kg;

The kg is a quantity of mass used to measure the amount of matter in a substance.

   Given mass = 5.25 x 10⁻¹³kg

The kilo-  is a prefix that denotes 10³

  therefore;

         1000g = 1kilogram

 the milli-  is a prefix that denotes 10⁻⁻³

       1000mg = 1g

Now that we know this, we can convert:

   5.25 x 10⁻¹³kg  x \frac{1000g}{1kg}  x \frac{1000mg}{1g}   =  5. 25 x 10⁻¹³ x 10⁶mg

      =  5. 25 x 10⁻⁷mg

learn more:

Conversion brainly.com/question/1548911

#learnwithBrainly

8 0
3 years ago
What is the process of changing a liquid into a gas?
WINSTONCH [101]
Evaporating
-----------
8 0
3 years ago
If the change in entropy of the surroundings for a process at 451 k and constant pressure is -326 j/k, what is the heat flow abs
Arisa [49]

If the change in entropy of the surroundings for a process at 451 k and constant pressure is -326 j/k, then heat flow absorbed (in kj) by the system is -147.026kJ.

<h3>What is entropy? </h3>

The entropy of particle is defined as how random it move. It shows the randomness of the system or may be disorders of the system. It is used to measure the unavailable energy for performing useful work.

Unit of entropy = J/K

<h3>Formula:</h3>

∆s = ∆Q/T

where,

∆s = change in entropy of the surrounding = -326J/K

∆Q = heat absorbed from surrounding

T = Temperature = 451K

∆Q = ∆s × T

∆Q = -326 × 451

∆Q = 147,026 J

∆Q = 147.026 kJ

Thus we find that the heat absorbed by the system is 147.026 kJ.

learn more about entropy:

brainly.com/question/14131507

#SPJ1

8 0
2 years ago
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