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love history [14]
3 years ago
15

Which of the following is not a weakness of the Rutherford model of the atom?

Chemistry
1 answer:
Harrizon [31]3 years ago
7 0
The answer to this is t<span>he atom is mostly empty space.</span>
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What is the pressure of 0.5 mol of nitrogen gas in a 5 L container at 203 K
Ivan

Answer:

1.67 atm.

Explanation:

  • We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = ??? atm).

V is the volume of the gas in L (V = 5.0 L).

n is the no. of moles of the gas in mol (n = 0.5 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 203 K).

∴ P = nRT/V = (0.5 mol)(0.0821 L.atm/mol.K)(203 K)/(5.0 L) = 1.67 atm.

3 0
3 years ago
Compare the solubility and melting points of ethane, ethanol, and ethanoic acid. Explain
OlgaM077 [116]
Ethane is a hydrocarbon with the chemical formula C2H6. It has a melting point of -297.9 F and a solubility of 56.8 mgL-1. Ethane is a gas at standard temperature and is not soluble in water.  Ethanol has the formula C2H5OH and is soluble in organic solvents and water. It has a melting point of -173.4 F. Ethanoic acid is soluble in water but not as soluble as ethanol. It has a melting point of 61.9 F. 
3 0
3 years ago
Given an initial cyclopropane concentration of 0.00560 m, calculate the concentration of cyclopropane that remains after 1.50 ho
maria [59]

<span>We can solve this problem by assuming that the decay of cyclopropane follows a 1st order rate of reaction. So that the equation for decay follows the expression:</span>

A = Ao e^(- k t) 

Where,

A = amount remaining at time t = unknown (what to solve for) <span>
Ao = amount at time zero = 0.00560 M </span><span>
<span>k = rate constant
t = time = 1.50 hours or 5400 s </span></span>

The rate constant should be given in the problem which I think you forgot to include. For the sake of calculation, I will assume a rate constant which I found in other sources:

k = 5.29× 10^–4 s–1                     (plug in the correct k value)

<span>Plugging in the values in the 1st equation:</span>

A = 0.00560 M * e^(-5.29 × 10^–4 s–1 * 5400 s )

A = 3.218 <span>× 10^–4 M           (simplify as necessary)</span>

8 0
2 years ago
Calculate the ratio of effusion rates of cl2 to f2 .
WINSTONCH [101]
<span>Answer: Graham's law of gaseous effusion states that the rate of effusion goes by the inverse root of the gas' molar mass. râšM = constant Therefore for two gases the ratio rates is given by: r1 / r2 = âš(M2 / M1) For Cl2 and F2: r(Cl2) / r(F2) = âš{(37.9968)/(70.906)} = 0.732 (to 3.s.f.)</span>
6 0
3 years ago
Why can liquid boil at a low pressure?
Paladinen [302]

Answer: I found this online. Hope it helps you.

Explanation:

This pressure is transmitted throughout the liquid and makes it more difficult for bubbles to form and for boiling to take place. If the pressure is reduced, the liquid requires less energy to change to a gaseous phase, and boiling occurs at a lower temperature.

7 0
3 years ago
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