Ernest Rutherford's gold-foil experiment showed which of the following?
Ernest Rutherford's gold-foil experiment showed which of the following? - 1

Answers

Answer 1
Answer: Ernest Rutherford's gold-foil experiment showed the density of atoms.
The experiment proved that most of an atom is empty space with a very small positively charged nucleus in the middle. 
So, from the given statements he following is true: 
Ernest Rutherford's gold-foil experiment showed the existence of a dense, positively charged center in an atom. 

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A real gas behaves least like an ideal gas under the conditions of(1) low temperature and low pressure
(2) low temperature and high pressure
(3) high temperature and low pressure
(4) high temperature and high pressure

Answers

The correct answer is option 3. A real gas behaves least like an ideal gas under the conditions of high temperature and low pressure. At this temperature and pressure, the molecules are close to each other and collisions or interactions are very likely to happen which is not an ideal gas.

What is the property of flowing very quickly in a liquid?

Answers

The answer is Fluidity.
Fluidity is the physical property that enables a substance to flow. Fluidity also refers to the state of being fluid. So basically, solid molecules has no fluidity.  It is also believed that the fluidity of water increases as the water gets warmer.

Which of the following has a positive enthalpy without a temperature change? condensing gas

evaporating a liquid

cooling a solid

cooling a liquid

Answers

A cooling solid because a cooling liquid would turn into a solid
 i think the answer is cooling a liquid

how much potassium bromide, in grams, should be added to water to prepare 0.50 of solution with molarity of 0.125M?

Answers

Volume = 0.50 L

Molar mass  KBr = 119.002 g/mol

Molarity = 0.125 M

Mass ( KBr) = ?

 mass = molarity * molar mass * volume

mass = 0.125 * 119.002 * 0.50

mass = 7.437625 g of KBr

hope this helps!

If the Kelvin temperature of a gas is tripled and the volume if doubled, the new pressure will be ___. A. 3/2 the original pressure B. 1/6 the original pressure C. 5 times the original pressure D. 2/3 the original pressurethe answer isnt b or c

Answers

The answer is 3/2 the original pressure. The concept behind this is the ideal gas law wherein it states that Pressure (P) times volume(V) is equal to number of moles (n) times ideal gas constant (R ) times temperature (T). Since you’re trying to calculate for the pressure you use P=nRT/V. Input 3 to Temperature and 2 to volume. You’ll get a pressure of 3/2 which is 3/2 times the original pressure. 

tert-Butyl alcohol is a solvent with a Kf of 9.10 ∘C/m and a freezing point of 25.5 ∘C. When 0.807 g of an unknown colorless liquid was dissolved in 11.6 g of tert-butyl alcohol, the solution froze at 15.3 ∘C.Which of the following is most likely the identity of this unknown liquid?tert-Butyl alcohol is a solvent with a of 9.10 and a freezing point of 25.5 . When 0.807 of an unknown colorless liquid was dissolved in 11.6 of tert-butyl alcohol, the solution froze at 15.3 .Which of the following is most likely the identity of this unknown liquid?ethylene glycol (molar mass = 62.07 g/mol)1-octanol (molar mass = 130.22 g/mol)glycerol (molar mass = 92.09 g/mol)2-pentanone (molar mass = 86.13 g/mol)1-butanol (molar mass = 74.12 g/mol)

Answers

Answer: ethylene glycol (molar mass = 62.07 g/mol)

Explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

\Delta T_f=k_f* m

or,

\Delta T_f=k_f* \frac{\text{mass of solute}}{\text{molar mass of solute}* \text{weight of solvent in kg}}

where,

T_f = change in freezing point

k_f = freezing point constant= 9.10^0C/m

m = molality

Given mass of solute = 0.807 g

Molar mass of solute=? g/mol

weight of solvent in kg= 11.6 g=0.0116 kg

\Delta T_f=T^(o)_f-T_f=25.5^0C-15.3^0C)=10.2^0C

10.2=9.10* \frac{0.807}{\text{molar mass of solute}* 0.0116kg}

{\text{molar mass of solute}}=62.07 g/mol

Thus the solute is ethylene glycol which has same molecular mass as calculated, i.e 62.07 g/mol.