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2 years ago

Which would most easily form negative one ions.

Chemistry

1 answer:

irinina [24]2 years ago

8 0

Answer:

Explanation:

chlorine (2,8,7) is a non metal with highest electronegativity. Hence, it is most likely to form a negative ion with charge −1.

I hope it helps you

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In your own words, explain how multiple approaches to a scientific investigation can be used to obtain the same result. Provide

aleksandr82 [10.1K]

Answer:B

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2 years ago

Mr. Smith is hyperglycemic with a blood glucose level of 300mg/ml of blood. Explain how homeostasis would regulate his glucose l

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Answer:

When the animal has eaten food and the blood glucose level in the body increases. The pancreas cells in the body detects the increase in the blood glucose which leads to increase the insulin hormone.

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Negative feedback is the way by which the body maintains homeostasis and maintains equilibrium in the body.

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2 years ago

In this experiment when both crests combine they produce a __________ spot.

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2 years ago

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3. How fast will her 1500 kg car accelerate if it produces 5000 N of force?

andrew-mc [135]

Answer:

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2 years ago

4. Magnesium and oxygen undergo a chemical reaction to

erastovalidia [21]

Answer:

About 16.1 grams of oxygen gas.

Explanation:

The reaction between magnesium and oxygen can be described by the equation:
$\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}$

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

Convert grams of MgO to moles of MgO.
Moles of MgO to moles of O₂
And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

$\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}$

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

$\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}$

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1 year ago