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3 years ago

What is the electron configuration of bromine whose atomic number is 35

Chemistry

1 answer:

Vanyuwa [196]3 years ago

5 0

1s22s22p63s23p64s23d104p5 is the answer

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A galvanic cell consists of a Al3 /Al half-cell and a standard hydrogen electrode. If the Al3 /Al half-cell standard cell functi

11111nata11111 [884]

Answer:

Explanation:

here is your answer:

3 0

3 years ago

Please help!!<br> I don’t understand

laiz [17]

<h3>Answer:</h3>

150 g Si

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Use Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 3.2 × 10²⁴ atoms Si

[Solve] grams Si

<u>Step 2: Identify Conversions</u>

Avogadro's Number

[PT] Molar Mass of Si - 28.09 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 3.2 \cdot 10^{24} \ atoms \ Si(\frac{1 \ mol \ Si}{6.022 \cdot 10^{23} \ atoms \ Si})(\frac{28.09 \ g \ Si}{1 \ mol \ Si})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 149.266 \ g \ Si$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. Instructed to round to 2 sig figs.</em>

149.266 g Si ≈ 150 g Si

4 0

3 years ago

Question 22 of 30

Black_prince [1.1K]

Answer:

C6H12O6+6O2--->6CO2+6H2O

Explanation:

So I went through all the answers and could not find the right one amongst. If I'm not wrong the reaction above is the reaction for respiration. The nearest answer is D but unfortunately the first reactant isn't in accordance with that which the question has given.

8 0

3 years ago

How many electrons would be found in an isotope of silicon 29

kenny6666 [7]

14 electrons15 neutrons

5 0

4 years ago

A 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. A

Kryger [21]

Answer:

$T_{2}=16,97^{\circ}C$

Explanation:

The specific heats of water and steel are

$Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}$

$Cp_{s}=0.49 \frac{KJ}{Kg^{\circ}C}$

Assuming that the water and steel are into an <em>adiabatic calorimeter</em> (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium $T_{2}_{w}= T_{2}_{s}$

An energy balance can be written as

$m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s}$

Replacing

$1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)$

Then, the temperature $T_{2}=16,97^{\circ}C$

8 0

3 years ago