Which substance is a gas at 20˚C and one atmosphere of pressure? A. C B. O3 C. Ca D. I2

Sodium is going in group 1, period 3. It contains A. One valence electron in its 3rd energy shell B three valence electron in it

topjm [15]

Answer:

The correct answer is option A, that is, one valence electron in its third energy shell and option C, that is, 11 electrons and 11 protons.

Explanation:

The outermost electrons and the ones that take part in the process of bonding are termed as valence electrons. The atomic number of sodium is 11, thus, it possesses 11 protons and the atoms are neutral so it suggests that sodium has 11 electrons. By electronic configuration, it can be seen that in sodium, two electrons are present in the first shell, 8 in the second, and only one electron in the third shell, that is, 2.8.1. The electron present in the third shell is the valence electron.

5 0

3 years ago

-5 - -2 plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz helppppppppppppppppppppppppppppppp meee

Zepler [3.9K]

The answer would be -3

4 0

2 years ago

Read 2 more answers

Using only the periodic table, determine the element that displays each given trend. Be sure to provide your reasoning.

RideAnS [48]

Choose the element that:

- is more reactive: oxygen or argon.

- has the larger atomic radius: francium or sodium.

- has the smaller atomic radius: scandium or selenium.

- has more ionization energy: bismuth or mercury.

- has less ionization energy: nitrogen or phosphorus.

- has more electronegativity: xenon or silver.

- has less electronegativity: titanium or hafnium.

Explanation:

Choose the element that is more reactive: oxygen or argon.

Argon have a complete electronic shell which makes him inert from chemically point of view while oxygen is able to form compounds with the majority of the elements.

Choose the element that has the larger atomic radius: francium or sodium.

As you go down in a group the atomic radius is increasing.

Choose the element that has the smaller atomic radius: scandium or selenium.

As a general trend the atomic radius of elements decrease in periods from right to left for this elements .

Choose the element that has more ionization energy: bismuth or mercury.

Mercury have a complete electronic shell (5d¹⁰) while bismuth have 3 electrons in the last shell (6p³). It is harder to extract an electron from mercury compared to bismuth.

Choose the element that has less ionization energy: nitrogen or phosphorus.

Ionization energy decrease as you go down in a group.

Choose the element that has more electronegativity: xenon or silver.

Silver have a higher electronegativity than Xenon. Xenon is an inert gas and it has a complete electronic shell which explains his inert chemical reactivity.

Choose the element that has less electronegativity: titanium or hafnium.

As you go down a grup electronegativity is decreasing.

Learn more about:

trends in periodic table:

brainly.com/question/2293240

brainly.com/question/4433739

brainly.com/question/3266672

#learnwithBrainly

3 0

3 years ago

Ammonia gas will react with oxygen gas to yield nitrogen monoxide gas and water vapor.

ruslelena [56]

Answer:

a) 0.168 moles O2

b) 9.50 grams O2

c) 0.01662 kg NO

d)88.9 %

Explanation:

Step 1: Data given

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation:

4NH3 + 5O2 → 4NO + 6H2O

a. How many moles of ammonia will react with 6.73g of oxygen?

Calculate moles of oxygen = mass O2/ molar mass O2

moles oxygen = 6.73 grams / 32.00 g/mol = 0.210 moles

Calculate moles of NH3

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.210 moles O2 we need 4/5 *0.210 = 0.168 moles O2

b. If 6.42g of water is produced, how many grams of oxygen gas reacted?

Calculate moles of H2O = 6.42 grams / 18.02 g/mol = 0.356 moles

Calculate moles of O2:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.356 moles H2O we'll need 5/6 * 0.356 = 0.297 moles O2

Calculate mass of O2 = moles O2 * molar mass O2

Mass O2 = 0.297 moles O2 * 32.00 g/mol = 9.50 grams O2

c. If the reaction uses up 9.43105 g of ammonia, how many kilograms of nitrogen monoxide will be formed?

Calculate moles of ammonia = 9.43105 grams / 17.03 g/mol =0.5538 moles

Calculate moles of NO:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.5538 moles of NH3 we'll have 0.5538 moles NO

Calculate mass of NO

Mass NO = 0.5538 moles * 30.01 g/mol = 16.62 grams = 0.01662 kg NO

d. When 2.51 g of ammonia react with 3.76 g of oxygen, 2.27 g of water vapor are produced. What is the percentage yield of water?

Calculate moles of NH3 = 2.51 grams / 17.03 g/mol = 0.147 moles

Calculate moles of O2 = 3.76 grams / 32 g/mol = 0.118 moles

Determine the limiting reactant

O2 is the limiting reactant, it will completely be consumed (0.118 moles)

NH3 is in excess. There will react 4/5 * 0.118 = 0.0944 moles

There will remain 0.147 - 0.0944 = 0.0526 moles

Calculate moles H2O: For 0.118 moles O2 we'll have 6/5 * 0.118 = 0.1416 moles H2O

Calculate mass H2O = 0.1416 moles * 18.02 g/mol = 2.552 grams H2O

Calculate % yield = (2.27/2.552)*100 % = 88.9 %

4 0

2 years ago

In order to prepare 50.0 mL of 0.100 M NaOH you will add _____ mL of 1.00 M NaOH to _____ mL of water

FinnZ [79.3K]

The question requires us to complete the sentence regarding the preparation of a more dilute NaOH solution (0.100 M, 50.0 mL) from a more concentrated NaOH solution (1.00 M).

Analyzing the blank spaces that we need to fill in the sentence, we can see that we must provide the volume of the more concentrated solution and the volume of water necessary to prepare the solution.

We can use the following equation to calculate the volume of more concentrated solution required:

$\begin{gathered} C_1\times V_1=C_2\times V_2 \\ V_1=\frac{C_2\times V_2}{C_1} \end{gathered}$

where C1 is the concentration of the initial solution (C1 = 1.00 M), V1 is the volume required of the inital solution (that we'll calculate), C2 is the concentration of the final solution (C2 = 0.100 M) and V2 is the volume of the final solution (V2 = 50.0 mL).

Applying the values given by the question to the equation above, we'll have:

$\begin{gathered} V_1=\frac{C_2\times V_2}{C_1} \\ V_1=\frac{0.100M_{}\times50.0mL_{}}{1.00M_{}}=5.00mL \end{gathered}$

Thus, we would need 5.00 mL of the more concentrated solution.

Since the volume of the final solution is 50.0 mL and it corresponds to the volume of initial solution + volume of water, we can calculate the volume of water necessary as:

$\begin{gathered} \text{final volume = volume of initial solution + volume of water} \\ 50.0mL=5.00mL\text{ + volume of water} \\ \text{volume of water = 45.0 mL} \end{gathered}$

Thus, we would need 45.0 mL of water to prepare the solution.

Therefore, we can complete the sentence given as:

"In order to prepare 50.0 mL of 0.100 M NaOH you will add 5.00 mL of 1.00 M NaOH to 45.0 mL of water"

5 0

1 year ago