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2 years ago

Using the periodic table to locate each element, write the electron configuration of(a) Ru;

1 answer:

4 0

Ruthenium is a chemical element with the symbol Ru and atomic number 44. The electron configuration of Ru is $5s14d^{7}$ .

<h3>How to write an electronic configuration?</h3>

Find the supplied element's atomic number on the periodic table and identify it.
The energy level and kind of orbital should be listed first, followed by the number of electrons in the orbital in superscript.
The Aufbau principle's diagonal rule for electron filling order in the various subshells is the simplest way to express the electronic configuration of any element.
The Aufbau rule, the Pauli-exclusion rule, and Hund's Rule are the three rules that must be followed when expressing the electron configuration in the orbital box diagram.

Ruthenium exists a chemical element with the symbol Ru and atomic number 44. The electron configuration of Ru is $5s14d^{7}$ .

To learn more about electronic configuration, refer to:

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14 billion years (roughly the age of the universe) to seconds. (Assume there are 365 days in a year.)

fenix001 [56]

Answer:

4.41 × 10¹⁷ Seconds

Explanation:

For this problem we will first find out the number of days in 14 billion years.

Remember that 1 billion is equal to 1,000,000,000 or 10⁹. So, 14 billion years will be written as 14,000,000,000 or 1.4 × 10¹⁰ years.

Therefore, In

1 year there are = 365 years

So, In

1.4 × 10¹⁰ years there will be = X years,

Solving for X,

X = 1.4 × 10¹⁰ years × 365 days / 1 year

X = 5.11 × 10¹² days

Secondly, as there are 24 hours per day and each hour has 60 minutes and each minute has 60 seconds, therefore, the number of seconds per day are calculated as,

Number of Sec/day = 24 × 60 × 60

Number of Sec/day = 86400 sec/day

Hence, If,

1 day has = 86400 seconds

then,

5.11 × 10¹² days will contain = X seconds

Solving for X,

X = 5.11 × 10¹² days × 86400 sec / 1 day

X = 4.41 × 10¹⁷ Seconds

6 0

4 years ago

What structure do bacteria cells have

g100num [7]

Answer:

They are clumped together and fluffy looking

Explanation:

I’ve seen what they looked like in a project during highschool we swabbed some books and door handles and saw which had the most bacteria. It was surprisingly the books

6 0

2 years ago

1. Phosphorous reacts with bromine to form phosphorous tribromide. If 35.0 grams of bromine

Anastaziya [24]

Answer:

70.6 %

Explanation:

First step, we define the reaction:

2P + 3Br₂ → 2PBr₃

We determine the moles of reactant:

35 g . 1mol / 159.8 g = 0.219 moles

We assume, the P is in excess, so the bromine is the limiting reagent.

3 moles of Br₂ can produce 2 moles of phophorous tribromide

Then, 0.219 moles may produce (0.219 . 2) /3 = 0.146 moles of PBr₃

We convert moles to mass:

0.146 mol . 270.67 g /mol = 39.5 g

That's the 100 % yield reaction, also called theoretical yield. The way to determine the % yield is:

(Yield produced / Thoeretical yield) . 100

(27.9 / 39.5) . 100 = 70.6 %

7 0

3 years ago

6.9 moles of oxygen gas in a fixed volume of 1.33 liters. One of the valves opened and there remains only 1.24 moles oxygen gas.

Romashka [77]

Answer:

The new volume of the gas remains the same. That is new volume of gas is 1.33 litres

Explanation:

This is because gases do not have a definite shape. They therefore take the shape of their containing vessels and hence their volumes are determined by the volume of the container.

For the question above even if some of the gas escapes, as long as there is gas present in the container, its volume remains the same, that is occupies the same space in the container

7 0

2 years ago

What is the predicted change in the boiling point of water when 1.50 g of

dezoksy [38]

Answer:

0.00735°C

Explanation:

By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water

⠀

$\textsf {While} \: \sf {\Delta T_b} \: \textsf{expression is used} \\ \textsf {for elevation of boiling point}$

⠀

The elevation in boiling point is a phenomenon in which there is increase in boiling point in solution, when the particular type of solute is added to pure solvent.

⠀

$\sf \large \underline{The \: formula \: to \: be \: used \: in \: this \: question \: is} \\ \boxed{T_b = i \times K_b \times m}$

⠀

Where 'i' is van't hoff factor which represents the ratio of observed osmotic pressure and the value to be expected.

and 'i' is 3 (as given in the question)

⠀

'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)

⠀

'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.

⠀

To find molality, we have to divide no. of moles of solute by weight of solution

⠀

While first we need to no. of moles

$\sf \implies no. \: of \: moles = \frac{weight \: of \: solute}{molar \: mass \: of \: solute} \\ \\ \implies \sf no. \: of \: moles = \frac{1.5}{208.23} \\ \\ \sf \implies no. \: of \: moles = 0.0072$

⠀

Now, we will find molality

⠀

$\sf \hookrightarrow molality = \frac{no.\: of \: moles}{weight \: of \: solution} \\ \\ \sf \hookrightarrow molality = \frac{0.072}{1.5} \\ \\ \sf \hookrightarrow molality = 0.048 \: mol {kg}^{ - 1}$

⠀

$\textsf{ \large{ \underline{Now substituting the required values}}}$

⠀

$\sf \longmapsto \Delta T_b = 3 \times 0.51 \times 0.0048 \\ \\ \\ \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}$

⠀

Henceforth, the change in boiling point is 0.00735°C.

7 0

2 years ago