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dolphi86 [110]
3 years ago
12

What is the most important use of an elements atomic number

Chemistry
1 answer:
Setler [38]3 years ago
8 0

Answer:

Atoms of each element contain a characteristic number of protons and electrons. The number of protons determines an element's atomic number and is used to distinguish one element from another.

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What is the mass of 2.80 grams of H2O
IrinaK [193]

The mass of 2.80 grams of h2o is 18.02 amu I believe

8 0
3 years ago
Could the Periodic Table be arranged differently?
coldgirl [10]

Answer:

No because it is stayed that way and you can't define them differently.

7 0
3 years ago
Read 2 more answers
50.0 g of NaNO3 are dissolved into enough water to make 250 mL of solution. What is the molarity of this solution?
d1i1m1o1n [39]

Answer:

2.35 M

Explanation:

Molarity is mol/L of solution. We have to convert the g to mol and the mL to L. G to mol uses the molar mass of the compound. The molar mass of NaNO₃ is 85.00g/mol.

50.0gNaNO3*\frac{1molNaNO3}{85.00gNaNO3} = 0.588molNaNO3

Then you have to convert mL to L.

250mL*\frac{1L}{1000mL} = 0.250L

Now divide the mol by the L.

\frac{0.588mol NaNO3}{0.250L} = 2.352 M

Round to the smallest number of significant figures = 2.35M

7 0
3 years ago
What is clch2ch2ch2oh when oxidized?
Maksim231197 [3]
The first step of the oxidation of a primary alcohol involves conversion to an aldehyde via the elimiination of a hydrogen molecule. Thus, ClCH2CH2CH2OH becomes ClCH2CH2COH, which is 3-chloropropanal.
5 0
3 years ago
Be sure to answer all parts. A concentration cell consists of two Sn/Sn2+ half-cells. The electrolyte in compartment A is 0.23 M
kolbaska11 [484]

Answer:

Part A. The half-cell B is the cathode and the half-cell A is the anode

Part B. 0.017V

Explanation:

Part A

The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).

So, the half-cell B is the cathode and the half-cell A is the anode.

Part B

By the Nersnt equation:

E°cell = E° - (0.0592/n)*log[anode]/[cathode]

Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.

E°cell = 0 - (0.0592/2)*log(0.23/0.87)

E°cell = 0.017V

3 0
3 years ago
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