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3 years ago

The periodic table of elements is organized by the number of what?

Chemistry

1 answer:

DENIUS [597]3 years ago

5 0

Protons, it was once organized by atomic mass but organizing by protons turned out to be better

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Steven had a sample of ethanol and wanted to see if it would boil at the temperature found in his textbook. His experiment yield

Natasha_Volkova [10]

Answer:

17.04%

Explanation:

Actual Value = 173.1

Measured Value = 143.6

Percent error is obtained using the equation;

Percent error = (Measured - Actual) / Actual ]* 100

Percent error = [ (143.6 - 173.1) / 173.1 ] * 100

The absolute value of (Measured - Actual) is taken,

Percent Error = [29.5 / 173.1 ] * 100

Percent Error = 0.1704 * 100 = 17.04%

4 0

3 years ago

A student balanced the chemical equation Mg + O2 →MgO by writing Mg + O2 → MgO2. Was the equation balanced correctly? Explain yo

Amanda [17]

Explanation:

Charges on both magnesium and oxygen is 2. Though opposite in sign, they have equal charges so, both of them will be cancelled by each other.

As a result, formula of magnesium oxide is MgO and not $MgO_{2}$ .

The student write the equation as $Mg + O_2 \rightarrow MgO_2$ , it is not correct.

Therefore, given equation will be balanced as follows.

$2Mg + O_{2} \rightarrow 2MgO$

Since, number of atoms on both reactant and product side are equal. Hence, this equation is completely balanced.

7 0

3 years ago

I will give brainliest and 76 points

katen-ka-za [31]

Answer:

number 5 is the answer

Explanation:

8 0

2 years ago

(4) Calculate the % of a compound that can be removed from liquid phase 1 by using ONE to FOUR extractions with a liquid phase 2

maksim [4K]

Answer:

One extraction: 50%

Two extractions: 75%

Three extractions: 87.5%

Four extractions: 93.75%

Explanation:

The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.

qⁿ = (V₁/(V₁ + KV₂))ⁿ

We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:

qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ = (V₁/(2V₁))ⁿ = (1/2)ⁿ

When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.

When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.

When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.

When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.

5 0

3 years ago

8. What volume (ml) of a 5.45 M lead nitrate solution must be diluted to 820.7 ml to make a

Ganezh [65]

212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

Answer:

Option A.

Explanation:

Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

$C_{1} V_{1} =C_{2} V_{2}$

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.

Then, $(5.45*V_{1} ) = (820.7*1.41)$

$V_{1}=\frac{820.7*1.41}{5.45}= 212.33 ml$

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

3 0

3 years ago