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beks73 [17]
3 years ago
15

Why is it important to know what weather fonts are

Chemistry
1 answer:
morpeh [17]3 years ago
4 0

Answer: In order to forecast the weather, they study high and low pressure systems and the boundaries between them which is why they are called weather fonts

Explanation:

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The following do not represent valid ground-state electron configurations for an atom either because they violate the Pauli excl
svetoff [14.1K]

Answer:

<em>For both cases the answer is C</em>

Explanation:  

We can see that the orbitals are not filled in the order of increasing energy and the Pauli exclusion principle is violated because it does not follow the correct order of the electron configuration; In the first exercise after the 2s2 orbital, the 2p2 orbital follows.

For the second exercise, you must start in order with level 1 and correctly filling each of the sublevels corresponding to each level until reaching level 7 and thus completing the desired number of electrons.

7 0
3 years ago
Read 2 more answers
A 20.00-mL sample of a weak base is titrated with 0.0568 M HCl. At the endpoint, it is found that 17.88 mL of titrant was used.
nata0808 [166]

Answer: 0.0508mL

Explanation: Using the basic formula that states: C acid * V acid = C base * V base. we have:0.568 * 17.88 = 20 * C base.

therefore concentration of the base is 1.0156/20 = 0.0508 mL

7 0
3 years ago
**ASAP**
jek_recluse [69]

the answer is C) . I hope this was helpful!

4 0
3 years ago
Read 2 more answers
MULTIPLE CHOICE
Flauer [41]

Answer:

its very simple ans we have 2 just multiply256

6 0
4 years ago
The gas phase decomposition of sulfuryl chloride at 600 K SO2Cl2(g)SO2(g) + Cl2(g) is first order in SO2Cl2. During one experime
krok68 [10]

Answer: The rate constant for the reaction is 3.96\times 10^{-3}min^{-1}

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample = 559 min

a = let initial amount of the reactant = 2.83\times 10^{-3}

a - x = amount left after decay process  = 3.06\times 10^{-4}

559min=\frac{2.303}{k}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}

k=\frac{2.303}{559}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}

k=3.96\times 10^{-3}min^{-1}

The rate constant for the reaction is 3.96\times 10^{-3}min^{-1}

6 0
4 years ago
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