2. What is the electron configuration of an electrically neutral atom of magnesium?

Be sure to answer all parts.

Liula [17]

A. The patch's area in square kilometers (km²) is 1.61×10⁻⁹ km²

B. The cost of the patch to the nearest cent is 734 cents

<h3>A. How to convert 16.1 cm² to square kilometers (km²)</h3>

We can convert 16.1 cm² to km² as illustrated below:

Conversion scale

1 cm² = 1×10⁻¹⁰ km²

Therefore,

16.1 cm² = 16.1 × 1×10⁻¹⁰

16.1 cm² = 1.61×10⁻⁹ km²

Thus, 16.1 cm² is equivalent to 1.61×10⁻⁹ km²

<h3>B. How to determine the cost in cent</h3>

We'll begin by converting 16.1 cm² to in². This can be obtained as illustrated below:

1 cm² = 0.155 in²

Therefore,

16.1 cm² = 16.1 × 0.155

16.1 cm² = 2.4955 in²

Finally, we shall the determine the cost in centas fo r llow:

Cost per in² = $2.94 = 294 cent
Cost of 2.4955 in² =?

1 in² = 294 cent

Therefore,

2.4955 in² = 2.4955 × 294

2.4955 in² = 734 cents

Thus, the cost of the patch is 734 cents

Learn more about conversion:

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6 0

2 years ago

A golf pro has 2100 J of kinetic energy and swings his driver which weighs .75 kg. What is the speed of his swing?

jeka57 [31]

Answer : The correct answer is 74.83 m/s .

The kinetic energy is energy possessed by any mass which is moving or have some speed . It is product of mass and velocity . It is expressed as :

$KE = \frac{1}{2} * m* v^2$

Where KE = kinetic energy in J or $kg\frac{m^2}{s^2}$ m = mass in Kg v = speed in m/s²

Unit of KE is Joules (J) .

Givne : KE = 2100 J mass = 0.75 kg v = ?

Plugging value in KE formula =>

$2100 J = \frac{1}{2} * 0.75 Kg * v^2$

$2100 J = 0.375 Kg * v^2$

Dividing both side by 0.375 kg =>

$\frac{2100 J}{0.375 Kg} = \frac{0.375 Kg}{0.375 Kg } * v^2$

$v^2 = 5600 \frac{m^2}{s^2}$

$v= 74.83 \frac{m}{s}$

8 0

3 years ago

Read 2 more answers

Is paint solution, suspension, colloid?

liraira [26]

Answer:

Collid

Explanation:

Answer might be that

5 0

3 years ago

Read 2 more answers

If a certain mass of mercury has a volume of 0.002 m2 at a temperature of 20 celsius, what will be the volume of 50 celsius?

SIZIF [17.4K]

Answer: 0.0022m3

Explanation:

Initial volume of mercury V1 = 0.002m3 (note that the unit of volume can be cm3, m3, dm3 or liters)

Initial temperature T1 = 20°C

Convert temperature in Celsius to Kelvin

( 20°C + 273°C = 293K)

Final temperature T2 = 50°C

( 50°C + 273°C = 323K)

Final volume V2 = ?

According to Charle's law, the volume of a fixed mass of a gas is directly proportional to the temperature.

Mathematically, Charles' Law is expressed as: V1/T1 = V2/T2

0.002/293 = V2/323

To get the value of V2, cross multiply

V2 = (323 x 0.002) / 293

V2 = 0.646 / 293

V2 = 0.0022 m3

Thus, the new volume of mercury will be 0.0022m3

3 0

3 years ago

Can someone help on five and six I am at a loss on how to solve it

Bumek [7]

Answer:

Q5.

a} Yes , All Cl2 get eliminated by adding 170 g of sodium thiosulfate .

b) There excess of Sodium thiosulfate by 2.637 g

Q6.

a) Mass = 162 g Sodium bicarbonate

b)Volume of Ammonia remain after the reaction = 0.086 x 24.8 = 1.60 L

Explanation:

Points to be considered :

There is a difference between STP and SATP :

STP = Standard Temperature and Pressure (273.15 K and 1 atm)

1 mole of gas at STP = 22.4 L

SATP = Standard Ambient Temperature and Pressure (293.15 K and 1 atm)

1 mole of gas at SATP = 24.8 L

$moles = \frac{Given\ mass}{Molar\ mass}$

Number of moles of gas at SATP :

$moles = \frac{Given\ Volume}{24.8L}$

Q5.

First, calculate the number of moles of Cl2 and thiosulfate present in the reaction :

Volume of Cl2 = 105 L

Moles of Cl2 =

$moles = \frac{Given\ Volume}{24.8}$

$moles = \frac{105}{24.8}$

Moles of Cl2 in Reaction medium = 4.2338 mole

Mass of Sodium thiosulfate = 170 g

Molar mass of thiosulfate = 158.11 g/mol (theoretical value)

$moles = \frac{170}{158.11}$

= 1.075

Moles of Sodium Thiosulfate in Reaction medium = 1.075 mole

To check whether the given moles of Cl2 and sodium thiosulfate satisfy theoretical values :

Consider the Given reaction and apply law of conservation of mass

$4Cl_{2}+Na_{2}S_{2}O_{3}+5H_{2}O\rightarrow 2NaHSO_{3}+8HCl$

$Na_{2}S_{2}O_{3}$ = sodium thiosulfate

This equation indicates ,

4 moles of Cl2 require = 1 mole of sodium thiosulfate

1 mole of Cl2 require =

$\frac{1}{4}$ of sodium thiosulfate = 0.25

4.2338 mole of Cl2 should need = 0.25 x 4.2338

= 1.058 mole of sodium thiosulfate

Required Thiosulfate = 1.058 mole

But,

Moles of Sodium Thiosulfate in Reaction medium = 1.075 mole

So , extra moles of Sodium Thiosulfate is present in the reaction by

= 1.075 - 1.0589 = 0.0166 mol

Molar mass of sodium thiosulfate = 158 .11 g/mol

$Mass =0.166\times 158.11$

= 2.637 g

a} .Yes , All Cl2 get eliminated by adding 170 g of sodium thiosulfate .

b) There excess of Sodium thiosulfate by 2.637 g

Q6.

Volume of ammonia = 50.0 L

Moles of Ammonia ,

$moles = \frac{Given\ Volume}{24.8L}$

$moles = \frac{50}{24.8L}$

= 2.016 moles

Moles of CO2 =

Mass of CO2 = 85.0 g

Molar mass = 44 g/mol

$moles = \frac{Given\ mass}{Molar\ mass}$

$moles = \frac{85.0}{44}$

= 1.932 mol of CO2

Now check the law of conservation of mass :

$NaCl + NH_{3} + CO_{2} +H_{2}O\rightarrow NaHCO_{3} +NH_{4}Cl$

According to above equation ,

1 mole of CO2 Needs = 1 mole of NH3

1.93 mol of CO2 need = (1 x 1.93) mol

1.93 mol of CO2 need = 1.93 mol of ammonia

Available ammonia = 2.016 mol

So Ammonia is in excess by:

= 2.016 - 1.93 mol

= 0.086 mol

Volume at SATP is calculated by

$V =mole\times 24.8$

Volume of Ammonia remain after the reaction = 0.086 x 24.8

= 2.1104 L

= 2.10 L

CO2 is the limiting reagent and governs the product formation :

Molar mass of NaHCO3 = 84.007 g/mol

1 mole of CO2 Needs = 1 mole of NaHCO3 = 84.007

1.93 mol of CO2 need = 1.93 x 84.007 mol

= 162.133 g of Sodium bicarbonate

= 162 g Sodium bicarbonate

Note : The answers are present in rounded figures .

6 0

3 years ago