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3 years ago

Can some one help me in this I hope this will be easy for you :)

Chemistry

1 answer:

kramer3 years ago

5 0

Answer:

c) i) Mg²⁺ ii) O²⁻

Explanation:

I don't know the answer to Q7. because you don't show the diagram

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The number of particles in one mole of a chemical substances is 602,200,000,000,000,000,000,000 particles. What is the correct w

crimeas [40]

The correct way to write this number in scientific notation : 6.022 x 10²³

<h3>Further explanation </h3>

Scientific Notation is a standard form of writing numbers used in mathematical operations

The rules:

1. Writing order : first digit, decimal point, next digit

2. Then followed by powers of 10 (the value can be +/-)

If the value is> 10 (the power of 10 is +), the decimal point is moved to the left, to the right of the first digit

If the value is <10 ((the power of 10 is -), the decimal point is moved to the right, to the right of the first digit

I certainly didn't count the details of the number of zeros listed, but only saw the answer choices.

And the right ones are: 6.022 x 10²³

3 0

3 years ago

Read 2 more answers

The pictures above show a gerbera daisy that has experienced a change called wilting. Which of the statements below best describ

Brilliant_brown [7]

Answer:

i just had this

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3 years ago

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____ [38]

Answer:

$m_{Mg}=30.8mgMg$

Explanation:

Hello,

Based on the given chemical reaction, as 31.2 mL of hydrogen are yielded, one computes its moles via the ideal gas equation under the stated conditions as shown below:

$n_{H_2}=\frac{PV}{RT}=\frac{754torr*\frac{1atm}{760torr}*0.0312L}{0.082 \frac{atm*L}{mol*K}*298.15K}=1.27x10^{-3}molH_2$

Now, since the relationship between hydrogen and magnesium is 1 to 1, one computes its milligrams by following the shown below proportional factor development:

$m_{Mg}=1.27x10^{-3}molH_2*\frac{1molMg}{1molH_2}*\frac{24.305gMg}{1molMg}*\frac{1000mgMg}{1gMg}\\m_{Mg}=30.8mgMg$

Best regards.

3 0

3 years ago

Calculate the radius of tantalum (Ta) atom, given that Ta has a BCC crystal structure, a density of 16.6 g/cm, and an atomic wei

Ivahew [28]

Answer:

The radius of tantalum (Ta) atom is $R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm$

Explanation:

From the Body-centered cubic (BBC) crystal structure we know that a unit cell length a and atomic radius R are related through

$a=\frac{4R}{\sqrt{3} }$

So the volume of the unit cell $V_{c}$ is

$V_{c}= a^3=(\frac{4R}{\sqrt{3} } )^3=\frac{64\sqrt{3}R^3}{9}$

We can compute the theoretical density ρ through the following relationship

$\rho=\frac{nA}{V_{c}N_{a}}$

where

n = number of atoms associated with each unit cell

A = atomic weight

$V_{c}$ = volume of the unit cell

$N_{a}$ = Avogadro’s number ( $6.023 \times 10^{23}$ atoms/mol)

From the information given:

A = 180.9 g/mol

ρ = 16.6 g/cm^3

Since the crystal structure is BCC, n, the number of atoms per unit cell, is 2.

We can use the theoretical density ρ to find the radio R as follows:

$\rho=\frac{nA}{V_{c}N_{a}}\\\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}$

Solving for R

$\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}\\\frac{64\sqrt{3}R^3}{9}=\frac{nA}{\rho N_{a}}\\R^3=\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}} \\R=\sqrt[3]{\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}$

Substitution for the various parameters into above equation yields

$R=\sqrt[3]{\frac{2\cdot 180.9}{16.6\cdot 6.023 \times 10^{23}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}\\R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm$

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4 years ago

Why is alchemy no longer accepted?

steposvetlana [31]

i think the answer is B. scientists used evidence, analysis, and experiments to disprove Alchemy. apologies if its incorrect.

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3 years ago