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3 years ago

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Name two elements that have properties similar to those of beryllium and have average atomic masses higher than 130

1 answer:

sertanlavr [38]3 years ago

3 0

In the periodic table, elements of the same group are characterized by having the same similar properties.
So, first we will check the elements that lie within the same group as <span>beryllium and then we will attempt to choose the elements with atomic mass higher than 130.

So, elements in the same group as </span>beryllium are: magnesium, calcium, strontium, barium and radium.
Among these elements, we will find that:
radium has atomic mass of 226 amu
barium has atomic mass of 137.327 amu

Based on this, the two elements would be barium and radium.

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Sulfur reacts with oxygen and creates two compounds. Compound A contains 1.34 g of sulfur for every 0.86 g of oxygen. Compound B

spayn [35]

Answer:

The mass ratio of oxygen rounded to the nearest whole no. 3 : 2

Explanation:

According to Law of Multiple proportion when two elements combine to make two or more different compounds, the mass ratio of the two element in the first compound, when divided by the mass ratio of the second compound , form a simple whole number ratio.

Compound A contains 1.34 g of sulfur for every 0.86 g of oxygen

$\frac{1.34}{0.86}= 1.5$

Compound B contains 11.63 g of sulfur for every 10.49 g of oxygen

$\frac{11.643}{10.49}= 1.0$

Ratio of oxygen in each compound

always put the larger number over the smaller number.

$\frac{CompoundA}{Compound B}=\frac{1.5}{1.0}=\frac{3}{2}$

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A chemist must prepare 300.0mL of nitric acid solution with a pH of 0.70 at 25°C. He will do this in three steps: Fill a 300.0mL

d1i1m1o1n [39]

<u>Answer:</u> The volume of concentrated solution required is 9.95 mL

<u>Explanation:</u>

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

pH = 0.70

Putting values in above equation, we get:

$0.70=-\log[H^+]$

$[H^+]=10^{-0.70}=0.199M$

1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.

Molarity of nitric acid = 0.199 M

To calculate the volume of the concentrated solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated nitric acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted nitric acid solution

We are given:

$M_1=7.0M\\V_1=?mL\\M_2=0.199M\\V_2=350mL$

Putting values in above equation, we get:

$7.0\times V_1=0.199\times 350.0\\\\V_1=\frac{0.199\times 350}{7.0}=9.95mL$

Hence, the volume of concentrated solution required is 9.95 mL

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Answer:

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