How many moles are in 9.83 x 1021 atoms of Na

Answer:

dog 2 because it is running 1m faster a second than dog 1. Dog 2 can run 1 meter farther in a second than the other dog.

Answer:

Age ≅ 7500 years

Explanation:

All radioactive decay is 1st order kinetics and described by the expression

A = A₀e^-kt => t = ln(A/A₀) / -k

k = 0.693 / t(half life) = (0.693 / 5730)yrs⁻¹ = 1.21 x 10⁻⁴ yrs⁻¹

t = Age = [ln(0.103/0.255) / - 1.21 x 10⁻⁴] yrs = 7500 years

Answer:

First, the protein may be a structural protein, contributing to the physical properties of cells or organisms. ... Second, the protein may be an enzyme that catalyzes one of the chemical reactions of the cell. Therefore, by coding for proteins, genes determine two important facets of biological structure and function.

Explanation:

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Answer:

There are approximately 366.25 sidereal days in a year so that the Earth spins 366.25 times with respect to distant stars in a year

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Explanation:

Answer:no

Explanation:

Answer:

Scandium is a transition metal

Explanation:

Answer:25,06 kJ of energy must be added to a 75 g block of ice.

ΔHfusion(H₂O) = 6,01 kJ/mol.

T(H₂O) = 0°C.

m(H₂O) = 75 g.

n(H₂O) = m(H₂O) ÷ M(H₂O).

n(H₂O) = 75 g ÷ 18 g/mol.

n(H₂O) = 4,17 mol.

Q = ΔHfusion(H₂O) · n(H₂O)

Q = 6,01 kJ/mol · 4,17 mol

Q = 25,06 kJ.

Explanation:

Answer:

5.450 mol Si₃N₄

Explanation:

Step 1: Write the balanced equation

3 Si + 2 N₂ ⇒ Si₃N₄

Step 2: Establish the theoretical molar ratio between the reactants

The theoretical molar ratio of Si to N₂ is 3:2 = 1.5:1.

Step 3: Establish the experimental molar ratio between the reactants

The experimental molar ratio of Si to N₂ is 16.35:11.26 = 1.45:1. Comparing both molar ratios, we can see that Si is the limiting reactant.

Step 4: Calculate the moles of Si₃N₄ produced from 16.35 moles of Si

The molar ratio of Si to Si₃N₄ is 3:1.

16.35 mol Si × 1 mol Si₃N₄/3 mol Si = 5.450 mol Si₃N₄

Answer:

Helium

Explanation:

Please tell me if it was right

Answer: the answer is a

Answer:

1 mole of ZnCl₂

Explanation:

Just from the stoichiometric equation/ balanced equation:

            Zn(s) + 2HCl(aq) → ZnCl₂(s) + H₂(g)

            1 mole   2 moles    1 mole       1 mole

Therefore: 2 moles of 2HCl produce 1 mole of ZnCl₂

Answer:

Explanation:

Because of the acid-base reaction, as sodium bicarbonate is introduced to the separatory funnel, the additional or unreacted HBr reacts vigorously to yield CO2 gas, which exits the separatory funnel together with any dissolved compound(s) in the ether layer. This is due to a wrong and incorrect selection of the solvent mixture and the addition of sodium bicarbonate to an acidic solution.

Nothing to do with it until it has leaked out of the separatory funnel. Even then, the student may separate the components from the remaining reaction mixture by washing the ether coating layer several times with brine water, then running it into a dry sodium sulfate bed and evaporating solvent ether under decreased pressure.

Answer:

[tex][SO_3]=0.25M[/tex]

Explanation:

Hello there!

In this case, since the integrated rate law for a second-order reaction is:

[tex][SO_3]=\frac{[SO_3]_0}{1+kt[SO_3]_0}[/tex]

Thus, we plug in the initial concentration, rate constant and elapsed time to obtain:

[tex][SO_3]=\frac{1.44M}{1+14.1M^{-1}s^{-1}*0.240s*1.44M}\\\\[/tex]

[tex][SO_3]=0.25M[/tex]

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Answer: The volume of tank is 17.5 L.

Explanation:

Given: Mass of methane = 173 g

Pressure = 15.1 atm

Temperature = 298 K

Molar mass of methane is 16.04 g/mol.

Therefore, moles of methane are calculated as follows.

[tex]No. of moles = \frac{mass}{molar mass}\\= \frac{173 g}{16.04 g/mol}\\= 10.78 mol[/tex]

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

[tex]PV = nRT\\15.1 atm \times V = 10.78 mol \times 0.0821 L atm/mol K \times 298 K\\V = \frac{10.78 mol \times 0.0821 L atm/mol K \times 298 K}{15.1 atm}\\= 17.5 L[/tex]

Thus, we can conclude that volume of the tank is 17.5 L.

Answer: [tex]2.49\ m^3[/tex]

Explanation:

Given

Mass of nitrogen present [tex]m=28\ g[/tex]

Temperature [tex]T=27^{\circ}C\equiv 300\ K[/tex]

Pressure [tex]P=785\ mm\ \text{of}\ Hg\ \text{or}\ 1.032\ atm[/tex]

The molar mass of Nitrogen [tex]M=28\ g/mol[/tex]

No of moles of nitrogen present

[tex]n=\dfrac{m}{M}\\\\n=\dfrac{28}{28}\\\\n=1[/tex]

Using [tex]PV=nRT[/tex]

[tex]\Rightarrow 1.032\times V=1\times 8.314\times 300\\\\\Rightarrow V=\dfrac{2494.2}{1.032}\\\\\Rightarrow V=2494.2\ L\ \text{or}\ 2.49\ m^3[/tex]

Answer:

A. Soil erosion will increase.

Option A is correct. Increase in wind speed will increase the rate of erosion.

The process of eroding, transferring, and depositing tiny particles as well as nutrients from the top layer of soil is known as soil erosion by wind.

In arid and semiarid regions, wind-driven soil erosion is a serious issue that impedes the environmental sustainability of animal husbandry and agriculture and threatens ecological security.

A wind that is blowing at 30 mph will erode at a rate that is more than 3 times faster than a wind that is blowing at 20 mph. As soil moisture rises, wind erosion declines. For instance, dry soil erodes approximately 1.3 times more quickly than soil with just enough moisture to support plant life.

Learn more about soil erosion;

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Answer:

Today, the average total production cost of “standard modulus” carbon fiber is in the range of $7-9 per pound.

Answer:

every 4 years we add an extra day, February 29th, to our calendars these extra days called leave days help synchronize our human created calendars with us orbit around the sun and the actual passing of our seasons

Domains eukarya  would this organism most likely fit.

The domain eukarya comprised of eukaryotes or organisms whose cells contain true nucleus.

It is  the largest of all groups in the classification of life. There are three domains:-

It is the domain of organism called eukaryotes. These are the organism who have a well defined nucleus and membrane bound organelles.

Hence, C) option is correct.

To know more about domain here

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Answer:

Preparation of Salts by The Action of An Acid Upon a Metal.

Preparation of Salts by Double Decomposition.

Preparation of Salts by Neutralization.

Using A Soluble Base (alkali)

(b). Using An Insoluble Base.

Preparation of Salts by the Action of An Acid on The Trioxocarbonate (IV) of A Metal.

Explanation:

Answer:

The final temperature is equal to 240 K or -33.15°C

Explanation:

Given that,

Initial temperature of the gas, T₁ = 47°C = 320 K

Initial pressure, P₁ = 140 kpa

Final pressure, P₂ = 105 kpa

We need to find the final temperature if the volume remains constant.  The relation between temperature and pressure is given by :

[tex]P\propto T[/tex]

or

[tex]\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}\\\\T_2=\dfrac{P_2T_1}{P_1}\\\\T_2=\dfrac{105\times 320}{140}\\\\T_2=240\ K\\\\T_2=-33.15^{\circ} C[/tex]

So, the final temperature is equal to 240 K or -33.15°C.

Answer:

False

Explanation:

Answer:

4.02g of Na2CO3⸱2H2O must be added completing the volume of the solution to 100mL

Explanation:

A 3%(w/v) solution contains 3g of solute (In this case, Na2CO3) in 100mL of solution.

Assuming we require 100mL of solution we must add 3g of Na2CO3. The reactant that is available is its dihydrate, with molar mass:

106g/mol + 2*MW H2O

106g/mol + 2*18g/mol = 142g/mol

That means the mass of Na2CO3.2H2O that must be added to prepare the solution is:

3g Na2CO3 * (142g/mol Na2CO3.2H2O / 106g/mol Na2CO3) =

Answer:

186.3g

Explanation:

4.5moles of K₂CO₃  is in 1000ml

? moles of K₂CO₃ is in 300 ml

(4.5 × 300)/ 1000 = 1.35 moles of K₂CO₃

1 mole of K₂CO₃ = (39 × 2) + 12 + (16 × 3) = 78 + 12 + 48 = 138g

1.35 moles of K₂CO₃  = ?

= (1.35 × 138)/1 = 186.3g

Answer: (1). There are  0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP.

(2). The density of gaseous arsine is 3.45 g/L.

Explanation:

1). At STP the pressure is 1 atm and temperature is 273.15 K. So, using the ideal gas equation number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

[tex]PV = nRT\\1 atm \times 0.372 L = n \times 0.0821 L atm/mol K \times 273.15 K\\n = 0.0165 mol[/tex]

2). As number of moles are also equal to mass of a substance divided by its molar mass.

So, number of moles of Arsine [tex](AsH_{3})[/tex] (molar mass = 77.95 g/mol) is as follows.

[tex]No. of moles = \frac{mass}{molar mass}\\0.0165 mol = \frac{mass}{77.95 g/mol}\\mass = 1.286 g[/tex]

Density is the mass of substance divided by its volume. Hence, density of arsine is calculated as follows.

[tex]Density = \frac{mass}{volume}\\= \frac{1.286 g}{0.372 L}\\= 3.45 g/L[/tex]

Thus, we can conclude that 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP and the density of gaseous arsine is 3.45 g/L.

Answer:

Warm front

Explanation:

A warm front forms when a warm air mass pushes into a cooler air mass, shown in the image to the right (A). Warm fronts often bring stormy weather as the warm air mass at the surface rises above the cool air mass, making clouds and storms. Warm fronts move more slowly than cold fronts because it is more difficult for the warm air to push the cold, dense air across the Earth's surface. Warm fronts often form on the east side of low-pressure systems where warmer air from the south is pushed north.

You will often see high clouds like cirrus, cirrostratus, and middle clouds like altostratus ahead of a warm front. These clouds form in the warm air that is high above the cool air. As the front passes over an area, the clouds become lower, and rain is likely. There can be thunderstorms around the warm front if the air is unstable.

On weather maps, the surface location of a warm front is represented by a solid red line with red, filled-in semicircles along it, like in the map on the right (B). The semicircles indicate the direction that the front is moving. They are on the side of the line where the front is moving. Notice on the map that temperatures at ground level are cooler in front of the front than behind it.

Answer:

An heater, Oven, sun, and fireplaces

Chose which ever you want