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2 years ago

The measure of central angle MNL is π radians, and the measure of the entire circle is 2π radians.

2 answers:

8 0

The ratio of the measure of the central angle to the entire circle measure is 1/2
The area of the entire circle is 36
The area of the sector is 18

chubhunter [2.5K]2 years ago

7 0

Answer: The ratio of the measure of the central angle to the entire circle measure is 1:2.

Explanation:

The measure of MNL angle = $\pi \text{ radians}$

The measure of the entire circle is = $2\pi \text{ radians}$

The ratio of the measure of the central angle to the entire circle measure is :

= $=\frac{Angle MNL}{\text{Angle of entire circle}}=\frac{\pi \text{ radians}}{2\pi \text{ radians}}=\frac{1}{2}$

Hence, the ratio of the measure of the central angle to the entire circle measure is 1:2.

Area of circle $=\pi r^2=\pi (unit)^2$

Area of sector = ratio of angle subtended by the sector to circle $\times$ area of circle

Area of sector = $\frac{1}{2}\times \pi =\frac{1}{2}\pi (unit)^2$

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What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

2 years ago

Small quantities of h2 gas can be collected by adding hcl to zn. a sample of 195 ml of h2 gas was collected over water at 25 c a

Umnica [9.8K]

15.4 milligrams The ideal gas law is PV = nRT where P = pressure of the gas V = volume of the gas n = number of moles of gas R = Ideal gas constant (8.3144598 L*kPa/(K*mol) ) T = absolute temperature. So let's determine how many moles of gas has been collected. Converting temperature from C to K 273.15 + 25 = 298.15 K Converting pressure from mmHg to kPa 753 mmHg * 0.133322387415 kPa/mmHg = 100.3917577 kPa Taking idea gas equation and solving for n PV = nRT PV/RT = n n = PV/RT Substituting known values n = PV/RT n = (100.3917577 kPa 0.195 L) / (8.3144598 L*kPa/(K*mol) 298.15 K) n = (19.57639275 L*kPa) / (2478.956189 L*kPa/(mol) ) n = 0.007897031 mol So we have a total of 0.007897031 moles of gas particles. Now let's get rid of that percentage that's water vapor. The percentage of water vapor is the vapor pressure of water divided by the total pressure. So 24/753 = 0.03187251 The portion of hydrogen is 1 minus the portion of water vapor. So 1 - 0.03187251 = 0.96812749 So the number of moles of hydrogen is 0.96812749 * 0.007897031 mol = 0.007645332 mol Now just multiple the number of moles by the molar mass of hydrogen gas. Start with the atomic weight. Atomic weight hydrogen = 1.00794 Molar mass H2 = 1.00794 * 2 = 2.01588 g/mol Mass H2 = 2.01588 g/mol * 0.007645332 mol = 0.015412073 g Rounding to 3 significant figures gives 0.0154 g = 15.4 mg

7 0

2 years ago

Read 2 more answers

A 150 W electric heater operates for 12.0 min to heat an ideal gas in a cylinder. During this time, the gas expands from 3.00 L

Brut [27]

Answer: The change in internal energy of the gas is 108.835 kJ

Explanation:

To calculate the work done for reversible expansion process, we use the equation:

$W=P\Delta V=-P(V_2-V_1)$

where,

W = work done

P = pressure = 1.03 atm

$V_1$ = initial volume = 3.00 L

$V_2$ = final volume = 11.0 L

Putting values in above equation, we get:

$W=-(1.03)\times (11.0-3.00)=8.24L.atm=834.9J=0.835kJ$ (Conversion factor: 1 L. atm = 101.325 J)

Calculating the heat from power:

$Q=P\times t$

where,

Q = heat required

P = power = 150 W

t = time = 12 min = 720 s (Conversion factor: 1 min = 60 s)

Putting values in above equation:

$Q=150\times 720=108000J=108kJ$

The equation for first law of thermodynamics follows:

$Q=dU+W$

where,

Q = total amount of heat required = 108 kJ

dU = Change in internal energy = ?

W = work done = -0.835 kJ

Putting values in above equation, we get:

$108kJ=dU+(-0.835)\\\\dU=(108+0.835)=108.835kJ$

Hence, the change in internal energy of the gas is 108.835 kJ

7 0

2 years ago

Rubbing alcohol contains 615g of isopropanol (C3H7OH) per liter (aqueous solution). Calculate the molality of this solution. Giv

faltersainse [42]

Answer:

Solution of isopropanol is 10.25 molal

Explanation:

615 g of isopropanol (C3H7OH) per liter

We gave the information that 615 g of solute (isopropanol) are contained in 1L of water. We need to find out the mass of solvent, so we use density.

Density of water 1g/mL → Density = Mass of water / 1000 mL of water

Notice we converted the L to mL

Mass of water = 1000 g (which is the same to say 1kg)

Molality are the moles of solute in 1kg of solvent, so let's convert the moles of isopropanol → 615 g . 1mol / 60g = 10.25 moles

Molality (mol/kg) = 10.25 moles / 1kg = 10.25 m

4 0

2 years ago

Combining aqueous solutions of bai2 and na2so4 affords a precipitate of baso4. Which ions are spectator ions in the reaction?

Marina86 [1]

Answer:

I⁻ (aq) and Na⁺ (aq)

Explanation:

We have the chemical reaction:

BaI₂ + Na₂SO₄ → BaSO₄ + 2 NaI

However if you want to determine the spectator ions you need to write the states of compounds:

(aq) - ions dissolved in water

(s) - solid

Ba²⁺ (aq) + 2 I⁻ (aq) + 2 Na⁺ (aq) + SO₄²⁻ (aq) → BaSO₄ (s) + 2 Na⁺ (aq) + 2 I⁻ (aq)

The ions which does not change the state and remains dissolved in the solution are spectator ions. For our chemical reaction we have the following spectator ions:

I⁻ (aq) and Na⁺ (aq)

3 0

2 years ago