What is one advantage to using this form of circuit? (AKS 10b) O A.

What is the difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm

Zanzabum

Answer:

The difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm is :

Difference in mass = 985.32 - 984.5 = 0.82 g

Explanation:

Part I :

$n =\frac{3.01\times 10^{24}}{6.022\times 10^{23}}$

n = 4.9983

n = 4.99 moles

(Note : You can also take n = 5 mole )

Molar mass of gold = 196.96 g/mole

This means, 1 mole of gold(Au) contain = 196.96 grams

So, 4.99 moles of gold contain = $5\times 196.96$ g

4.99 moles of gold contain = 984.8 g

Mass of ${3.01\times 10^{24}}$ atoms of gold = 984.5 g

Part II :

Density of Gold = $19.32 g/cm^{3}$

Volume of the cuboid = $length\times breadth\times height$

Volume of the gold bar = $6.00\times 4.25\times 2.00$

Volume of the gold bar = 51 $cm^{3}$

Using formula,

$Density = \frac{mass}{Volume}$

$Mass = Density\times Volume$

$Mass = 19.32 \times 51$

Mass = 985.32 g

So, A gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm has mass of 985.32 g

Difference in mass = 985.32 - 984.5 = 0.82 g

3 0

3 years ago

Someone help 6th grade science i will give brainliest

patriot [66]

I believe the answer is C

3 0

2 years ago

Read 2 more answers

How did you use QPOE?

Jlenok [28]

QPOE Files

The x-ray data are stored in QPOE files (Quick Position-Ordered Events, *.qp) rather than image arrays. These are lists of photons identified by several quantities, including the position on the detector, pulse height, and arrival time. Note that, unlike IRAF images, QPOE files have no associated header file, and are always stored in the current directory, unless explicitly specified otherwise. Non-PROS IRAF tasks can also access QPOE data files in place of image arrays.

8 0

2 years ago

Solution of the Schrödinger wave equation for the hydrogen atom results in a set of functions (orbitals) that describe the behav

kvv77 [185]

Answer :

'n' specifies → (B) The energy and average distance from the nucleus.

'l' specifies → (C) The subshell orbital shape.

'ml' specifies → (A) The orbital orientation.

Explanation :

Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as $m_l$ . The value of this quantum number ranges from $(-l\text{ to }+l)$ . When l = 2, the value of

Spin Quantum number : It describes the direction of electron spin. This is represented as $m_s$ . The value of this is $+\frac{1}{2}$ for upward spin and $-\frac{1}{2}$ for downward spin.

As per question we conclude that,

'n' specifies → The energy and average distance from the nucleus.

'l' specifies → The subshell orbital shape.

'ml' specifies → The orbital orientation.

5 0

3 years ago

1) For the following reaction, 8.44 grams of carbon (graphite) are allowed to react with 9.63 grams of oxygen gas.C(graphite)(s)

Yuki888 [10]

Answer:

The answers to the question are

(a) 13.24 g

(b) (O₂)

(c) 4.8252 g

(2) 0.662 M/L

Explanation:

(a) To solve the question we write the equation as follows

C + O₂ → CO₂

That is one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide

number of moles of graphite = 8.44/12 = 0.703 moles

number of moles of oxygen = 9.63/32 = 0.3009

However since one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide, therefore, 0.3009 moles of oxygen will react with 0.3009 moles of carbon to fore 0.3009 moles of CO₂

The maximum mass of carbon dioxide that can be formed = mass = moles × molar mass

= 0.3009×44 = 13.24 g

(b) The formula for the limiting reagent (O₂)

Finding the limting reagent is by checking the mole balance of the reactants available to the moles specified in th stoichiometry of the reaction and selecting the reagent with the list number of moles

(c) The mass of excess reagent = 0.703 moles - 0.3009 moles = 0.4021 moles

mass of excess reagent = 0.4021 × 12 = 4.8252 g

(2) The molarity is given by number of moles per liter of solution, therefore

molar mass of mgnesium iodide = 278.1139 g/mol, number of moles of magnesium iodide in 23 g = 23g/ 278.1139 g/mol= 8.3 × 10⁻² M

Therefore the moles in 125 mL = (8.3 × 10⁻² M)/(125 mL) = (8.3 × 10⁻² M)/(0.125 L) = 0.662 M/L

3 0

3 years ago