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xxTIMURxx [149]
3 years ago
11

Plz answer quick

Chemistry
2 answers:
SCORPION-xisa [38]3 years ago
8 0
It’s baron I got it right on edg
meriva3 years ago
3 0

Answer:

It is boron

Explanation:

If there are more questions like this, search something like this on Google

"Does ____ have three valence electrons?"

It should be right there

Hope this helps!

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Will mark brainliest if correct
Sergeeva-Olga [200]

Answer:

a and d

Explanation:

4 0
2 years ago
What is the mass of potassium chloride when 6.75 g of potassium reacts with an excess of chlorine gas? the balanced chemical equ
lorasvet [3.4K]
The balanced equation for the above reaction is;
2K + Cl₂ ---> 2KCl
Stoichiomtery of K to KCl is 2:2
Potassium is the limiting reactant which is fully consumed in the reaction. The amount of product formed depends on amount of limits reactant present.
Number of moles of K reacted - 6.75 g/ 39 g/mol = 0.17 mol
Therefore number of KCl moles formed - 0.17 mol
Mass of KCl formed - 0.17 mol x 74.5 g/mol = 12.67 g 
6 0
2 years ago
Find the initial concentration of the weak acid or base in each of the following aqueous solutions: (a) a solution of HClO with
Luda [366]

Answer:

a) 0.021 M

b) 0.019 M

Explanation:

To do this, you need to calculate the concentration of ions in solution with the given value of pH for each solution, then, write the chemical equation for both solutions, Set an ICE chart, use the value of Ka and Kb reported for both solutions, and solve for the initial concentration.

This is the general procedure to do it, now let's do it by parts.

<em><u>a) Concentration of HClO pH = 4.6</u></em>

With the given pH, we use the following expression:

pH = -log[H₃O⁺]      From here, we solve for [H₃O⁺]

[H₃O⁺] = 10^(-pH)   (1)

Let's calculate first the hydronium concentration:

[H₃O⁺] = 10^(-4.6) = 2.51x10⁻⁵ M

This value indicates the equilibrium concentration of this ion in solution. Now, to know the initial concentration of the acid, we need to do an ICE chart and write the chemical equation. This is an acid - base reaction, so we need the value of Ka of the acid.

         HClO + H₂O <---------> H₃O⁺ + ClO⁻       Ka = 3x10⁻⁸

I:            Y                                 0          0

C:          -x                                +x         +x

E:           Y - x                            x          x

With this chart, we need to write the expression for Ka which is:

Ka = [H₃O⁺] * [ClO⁻] / [HClO] = x² / Y-x

But we already know the concentration of [H₃O⁺], which is the same for [ClO⁻], and the value of Ka, so all we have to do is replace the values in the above expression and solve for Y:

3x10⁻⁸ = (2.51x10⁻⁵)² / Y - 2.51x10⁻⁵

We can round to Y because "x" is a very small value as it's value of Ka so:

3x10⁻⁸ = (2.51x10⁻⁵)²/Y

Y = (2.51x10⁻⁵)²/3x10⁻⁸

<h2><em>Y = [HClO] = 0.021 M</em></h2>

<em>And this is the initial concentration of the acid.</em>

<u><em>b) Solution of hidrazine pH = 10.2</em></u>

We do the same procedure as part a) with the difference that instead of using Ka , we use Kb and concentration of [OH⁻]. The Kb for hydrazine is 1.3x10⁻⁶

Let's calculate the [OH⁻]:

pOH = 14 - pH

pOH = 14 - 10.2 = 3.8

[OH⁻] = 10^(-3.8) = 1.58x10⁻⁴ M

The chemical equation:

          N₂H₄ + H₂O <---------> N₂H₅⁺ + OH⁻    Kb = 1.3x10⁻⁶

I:            Y                                  0           0

C:          -x                                +x           +x

E:         Y-x                                 x           x

Kb = x²/(Y-x)

1.3x10⁻⁶ = (1.58x10⁻⁴)²/Y

Y = (1.58x10⁻⁴)²/1.3x10⁻⁶

<h2><em><u>Y = [OH⁻] = 0.019 M</u></em></h2>

And this is the initial concentration of hydrazine

4 0
2 years ago
The air that we breathe is actually a mixture of several gases. of those involved, which is the third most abundant component of
labwork [276]
Co2 oxgen and nitrogen
4 0
3 years ago
A 57.07 g sample of a substance is initially at 24.3°C. After absorbing of 2911 J of heat, the temperature of the substance is 1
icang [17]

Answer:

Approximately 0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}.

Explanation:

The specific heat of a material is the amount of energy required to increase unit mass (one gram) of this material by unit temperature (one degree Celsius.)

Calculate the increase in the temperature of this sample:

\Delta T = (116.9 - 24.3)\; \rm ^\circ\! C= 92.6\; \rm ^\circ\! C.

The energy that this sample absorbed should be proportional the increase in its temperature (assuming that no phase change is involved.)

It took 2911\; \rm J of energy to raise the temperature of this sample by \Delta T = 92.6\; \rm ^\circ\! C. Therefore, raising the temperature of this sample by 1\; \rm ^\circ\! C (unit temperature) would take only \displaystyle \frac{1}{92.6} as much energy. That corresponds to approximately 31.436\; \rm J of energy.

On the other hand, the energy required to raise the temperature of this material by 1\; \rm ^\circ\! C is proportional to the mass of the sample (also assuming no phase change.)

It took approximately 31.436\; \rm J of energy to raise the temperature of 57.07\; \rm g of this material by 1\; \rm ^\circ C. Therefore, it would take only \displaystyle \frac{1}{57.07} as much energy to raise the temperature of 1\; \rm g (unit mass) of this material by 1\; \rm ^\circ \! C\!. That corresponds to approximately 0.551\; \rm J of energy.

In other words, it takes approximately 0.551\; \rm J to raise 1\; \rm g (unit mass) of this material by 1\; \rm ^\circ \! C. Therefore, by definition, the specific heat of this material would be approximately 0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}.

8 0
2 years ago
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