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3 years ago

In the gold foil experiment, what was observed

1 answer:

8 0

Rutherford's Gold Foil Experiment proved the existence of a small massive center to atoms, which would later be known as the nucleus of an atom. Ernest Rutherford, Hans Geiger and Ernest Marsden carried out their Gold Foil Experiment to observe the effect of alpha particles on matter.

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Calculate the volume in ml of 0.200 M na2co3 needed to produce 2.00 g of caco3 there is an excess of cacl2

UkoKoshka [18]

Answer:

$V=100mL$

Explanation:

Hello.

In this case, since the chemical reaction is:

$Na_2CO_3+CaCl_2\rightarrow CaCO_3+2NaCl$

We next compute the moles of sodium carbonate from the 2.00 grams of calcium carbonate via their 1:1 mole ratio in the chemical reaction:

$n_{Na_2CO_3}=2.00gCaCO_3*\frac{1molCaCO_3}{100.09gCaCO_3}*\frac{1molNa_2CO_3}{1molCaCO_3} \\\\n_{Na_2CO_3}=0.0200molNa_2CO_3$

Thus, by knowing the molarity, we compute the volume:

$M=\frac{n}{V}\\ \\V=\frac{n}{M}=\frac{0.0200mol}{0.200mol/L}\\ \\V=0.100L*\frac{1000mL}{1L}\\ \\V=100mL$

Best regards.

8 0

3 years ago

14. Which of the following is true of the nucleus?

velikii [3]

A)
Because it does not contain exactly all the DNA in the cell and some cells don't have a nucleus٫ such as the red blood cells.

5 0

3 years ago

Which of the following mixtures could not be separated by filtration

____ [38]

The answer to this question would be Salty water(i think its the most likely to go through) because salt water is a homogeneous mixture so it would go through.

4 0

3 years ago

List four reasons why the temperature of a flame is lower than the adiabatic flame temperature."

Nutka1998 [239]

Answer:

substances is never entirely a one way

Explanation:

no possibility of obtaining complete combustion at hig

temperature.

always loss of heat from the temperature

its temperature is lower than the constant volume process because some of the energy is utilized to change the volume of the system

7 0

3 years ago

in carbon tetrachloride proceeds as follows: 2N2O5→4NO2+O2. The rate law is first order in N2O5. At 64 ∘C the rate constant is 4

Nesterboy [21]

Answer:

Part A

$rate= 4.82*10^{-3}s^{-1} * [N2O5]$

Part B

$rate= 1.35*10^{-4}Ms^{-1}$

Explanation:

Part A

<em>The rate law is the equation that relates the rate of the reaction, the kinetic constant and the concentration of the reactant or reactants.</em>

For the given chemical reaction we can write a general expression for the rate law as follows:

$rate= k * [N2O5]^{x}$

where k is the rate constant and x is the order of the reaction with respect of N2O5 concentration. Particularly, <em>a first order reaction kinetics indicate that the rate of the reaction is directly proportional to the concentration of only one reactant</em>. Then x must be 1.

Replacing the value of the rate constant given in the text we can arrive to the following expression for the rate law:

$rate= 4.82*10^{-3}s^{-1} * [N2O5]$

Part B

Replacing the value of the concentration of N2O5 given, we can get the rate of reaction:

$rate= 4.82*10^{-3}s^{-1} *2.80*10^{-2}M$

$rate= 1.35*10^{-4}Ms^{-1}$

8 0

3 years ago