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ElenaW [278]
4 years ago
15

Only someone who truly knows and understands this, please tell me if the electron configurations are right :

Chemistry
1 answer:
Setler79 [48]4 years ago
4 0
Yes is 4s^23d^7=cobalt
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Plz need help hw plz help struggling
Mademuasel [1]
All you have to do is research more about this scientist´s theory. Just google: Why wasn´t Wegner´s hypothesis accepted?
Trust me, the answers are there.<span>
</span>Next time just research it.
Wegner´s idea seemed so ridiculous at the time that he was ridiculed by other scientists. The problem was that to his colleagues there seemed no plausible mechanism for how the continents could move through the oceans.
6 0
4 years ago
Which are true of the Celsius scale?
MAVERICK [17]

Answer:

B:) The freezing point of water is 32 degrees.

Explanation:

Hope it helps! :)

7 0
2 years ago
Convert .4076grams into moles<br> Element is copper
Artist 52 [7]

Answer:

0.00642mole

Explanation:

Molar Mass of Cu = 63.5g/mol

Mass of Cu from the question = 0.4076g

Number of mole =?

Number of mole = Mass /Molar Mass

Number of mole of Cu = 0.4076/63.5 = 0.00642mole

8 0
4 years ago
Consider the reaction of glucose with oxygen: C6H12O6(s) + 6O2(g) Right arrow. 6CO2(g) + 6H2O(l)
lidiya [134]

Answer:

Subtract the sum of the heats of formation of the reactants from that of the products to determine delta H: delta H = –110.53 kJ/mol – (–285.83 kJ/mol) = 175.3 kJ.

Explanation:

Step 1: Set Up the Equation. Arrange your given ΔHf and ΔH values according to the following equation: ΔH = ΔHf (products) - ΔHf (reactants). ...

Step 2: Solve the Equation. Solve your equation for ΔHf. ...

Step 3: Validate the Sign. Adjust your ΔHf value's sign depending on whether it is for a product or a reactant.

3 0
3 years ago
Can someone help me answer this homework problem. this is my homework.
sleet_krkn [62]

Answer:

12

Explanation:

The coefficient of the CO2 shows the amount of CO2 needed in the reaction so to react with 11 water, 12 molecules of carbon dioxide is needed.

8 0
3 years ago
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