142Pm > 142 Nd + what is the answer to this equation

What is electron configuration of oxygen in its excited state

timofeeve [1]

Answer:

$1 {s}^{2} 2 {s}^{2} 2 {p}^{4}$

OR

$2 : 6$

3 0

3 years ago

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Given 50.0 grams of Strontium, find grams of Phosphorus required for complete reaction

Sveta_85 [38]

Answer:

Mass = 11.78 g of P₄

Explanation:

The balance chemical equation is as follow:

6 Sr + P4 → 2 Sr₃P₂

Step 1: Calculate moles of Sr as;

Moles = Mass / M/Mass

Moles = 50.0 g / 87.62 g/mol

Moles = 0.570 moles

Step 2: Find moles of P₄ as;

According to equation,

6 moles of Sr reacted with = 1 mole of P₄

So,

0.570 moles of Sr will react with = X moles of P₄

Solving for X,

X = 1 mol × 0.570 mol / 6 mol

X = 0.0952 mol of P₄

Step 3: Calculate mass of P₄ as,

Mass = Moles × M.Mass

Mass = 0.0952 mol × 123.89 g/mol

Mass = 11.78 g of P₄

8 0

3 years ago

A neutral solution has a pH of: a. 1 b. 5 c. 7 d. 14

zavuch27 [327]

The answer is 7 my guy

3 0

3 years ago

What did Ernest Rutherford discover? How was his model different from Thomson’s?

s2008m [1.1K]

Answer:

thomson developed the chocolate chip method which was the identification of the electrons in the core of an atom. Rutherford discovered that the core was only positive and that the electrons were floating outside of the core.

Explanation:

4 0

3 years ago

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

Answer: The empirical formula for the given compound is $C_3H_6O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

3 0

3 years ago