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2 years ago

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A sample that weighs 107.75 g is a mixture of 30% helium atoms and 70% krypton atoms. How many particles are present in the samp

le?

1 answer:

r-ruslan [8.4K]2 years ago

5 0

<span>A sample that weighs 107.75 g is a mixture of 30% helium atoms and 70% krypton atoms. How many particles are present in the sample? </span>3.59

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Jared sprayed a pesticide on his garden three times. The first time, many insects died. The second time fewer insects died. The

olga_2 [115]

Answer:

C. The species became more resistant to the insecticide.

Explanation:

The more of something you receive, the more immune you are to it. For example, every year, many people get a flu shot. Flu shots contain small amounts of the flu, and it is inserted into your blood stream so that you may become more immune to the disease when it comes around. The same happens for many living things, including insects.

5 0

2 years ago

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A scientist measures the speed of sound in a monatomic gas to be 449 m/s at 20∘C. What is the molar mass of this gas?

Ivan

Answer:

The molar mass of the gas is 36.25 g/mol.

Explanation:

To solve this problem, we can use the mathematical relation:

ν = $\sqrt{3RT/M}$

Where, ν is the speed of light in a gas <em>(ν = 449 m/s)</em>,

R is the universal gas constant <em>(R = 8.314 J/mol.K)</em>,

T is the temperature of the gas in Kelvin <em>(T = 20 °C + 273 = 293 K)</em>,

M is the molar mass of the gas in <em>(Kg/mol)</em>.

ν = $\sqrt{3RT/M}$

(449 m/s) = √ (3(8.314 J/mol.K) (293 K) / M,

<em>by squaring the two sides:</em>

(449 m/s)² = (3 (8.314 J/mol.K) (293 K)) / M,

∴ M = (3 (8.314 J/mol.K) (293 K) / (449 m/s)² = 7308.006 / 201601 = 0.03625 Kg/mol.

<em>∴ The molar mass of the gas is 36.25 g/mol.</em>

7 0

2 years ago

The atomic mass of carbon is 12.01, sodium is 22.99, and oxygen is 16.00. What is the molar mass of sodium oxalate (Na2C2O4)?

nadya68 [22]

Answer:

134g/mol

Explanation:

4 0

2 years ago

A sodium ion, Na+, with a charge of 1.6×10−19C and a chloride ion, Cl− , with charge of −1.6×10−19C, are separated by a distance

sasho [114]

Answer:

$W\geq 2.1x10^{-19}J$

Explanation:

Due to Coulomb´s law electric force can be described by the formula $F=K\frac{q_{1}.q_{2}}{r^{2}}$ , where K is the Coulomb´s constant ( $9x10^{9} N\frac{m^{2} }{C^{2} }$ ), $q_{1}$ = Charge 1 (Na+ in this case), $q_{2}$ is the charge 2 (Cl-) and r is the distance between both charges.

Work made by a force is W=F.d and total work produced is the change in energy between final and initial state. this is $W=W_{f} -W_{i}$ .

so we have $W=W_{f} -W_{i} =(K\frac{q_{(Na+)}q_{(Cl-)}rf}{r_{f} ^{2}})-(K\frac{q_{(Na+)}q_{(Cl-)}ri}{r_{i} ^{2}})=Kq_{(Na+)}q_{(Cl-)[\frac{1}{{r_{f}}} -\frac{1}{{r_{i}}}]$

Given that ri= 1.1nm= $1.1x10^{-9}m$ and rf= infinite distance

$W=(9x10^{9})(1.6x10^{-19})(-1.6x10^{-19})[\frac{1}{\alpha }-\frac{1}{(1.1x10^{-9})}]=2.1x10^{-19}J$

6 0

2 years ago

A Gas OCCUPIES 525ML AT A PRESSURE OF 85.0 kPa WHAT WOULD THE VOLUME OF THE GAS BE AT THE PRESSURE OF 65.0 kPa

german

Boyle's law of ideal gas: This law states that the volume of a gas is inversely proportional to its pressure at a constant temperature. Acc to this law we can write the relation of pressure and volume as:

$PV=Constant$

That means:

$P_{1}V_{1}=P_{2}V_{2}$

From that equation we can calculate Volume of gas at a certain pressure:

P₁=Initial pressure

V₁=Initial volume

P₂=Final pressure

V₂= Final volume

Here P₁, initial pressure is given as 85.0 kPa

V₁, initial volume is given as 525 mL

P₂, final pressure is 65.0 kPa

$P_{1}V_{1}=P_{2}V_{2}$

so,

V_{2}=85\times 525\div 65

=686 mL

Volume of gas will be 686 mL.

8 0

2 years ago

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