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Natasha2012 [34]
3 years ago
12

The sun rotates on it axis once every: 28 hours 120 days 25 days 9 hours

Chemistry
1 answer:
olganol [36]3 years ago
8 0

Answer:

Itś about 27 days so just pick 28!

Explanation:

Hope this helped!

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Rainbow [258]

Answer:

Chicken nuggets

hope it helps have a nice day

6 0
2 years ago
Read 2 more answers
Which equation represents a neutralization reaction?
Harrizon [31]

Answer: HCI + KOH → KCI + H20

Explanation:

HCI(aq) + KOH(aq) → KCI(aq) + H20(l)

Acid + base → Salt + Water.

The above is a neutralization reaction in which an acid, aqeous HCl reacts completely with an appropriate amount of a base, aqueous KOH to produce salt, aqueous KCl and water, liquid H2O only.

This is a neutralization reaction since, the hydrogen ion, H+, from the HCl is neutralized by the hydroxide ion, OH-, from the KOH to form the water molecule, H2O and salt, KCl only.

3 0
3 years ago
Elements ending in the electron configurations ns^1 are highly reactive metals . What family does these elements belong to ?
myrzilka [38]

Answer:

<h3>A . Alkali metals</h3>

Explanation:

The highlighted elements of the periodic table belong to the alkali metal element family. The alkali metals are recognized as a group and family of elements. These elements are metals. Sodium and potassium are examples of elements in this family.

hope this helps

6 0
3 years ago
The pH scale is called the logarithmic scale what does this mean?
Ratling [72]
A logarithmic scale is a nonlinear scale used when there is a large range of quantities
6 0
3 years ago
Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin
STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of KI and AgNO_3.

\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}

Molar mass of KI = 166 g/mole

\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole

and,

\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

KI+AgNO_3\rightarrow KNO_3+AgI

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of AgNO_3

So, 0.0178 mole of KI react with 0.0178 mole of AgNO_3

From this we conclude that, AgNO_3 is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgI

From the reaction, we conclude that

As, 1 mole of KI react to give 1 mole of AgI

So, 0.0178 moles of KI react to give 0.0178 moles of AgI

Thus,

Moles of AgI = Moles of I^- anion = Moles of Ag^+ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}

\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

3 0
3 years ago
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