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3 years ago

Which model of the atom was the first to contain a nucleus?

1 answer:

4 0

Ernest Rutherford's gold foil experiment was when he shot positively charged particles at a thin layer of gold foil. He concluded that there was a core that contained most of the atom's mass, the core is the nucleus

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Elements with great electron affinities (large negative values) often have

Svet_ta [14]

Answer:

3. small sizes and high ionization energies.

Explanation:

Hello,

Ionization energies are always related with the formation of positive ions. On the other hand, electron affinities are the negative ion equivalent, and their use is almost always confined to elements in groups 6 and 7 of the Periodic Table

Small sizes and high ionization energies turn out into great electron affinities since it is easier for an electron to be added to an atom if it is small and it has a high ionization energy to promote the aforesaid addition.

Best regards.

7 0

3 years ago

If a chemical reaction such as photosynthesis begins with 6 atoms of carbon C , how many atoms of carbon C should be in the prod

pochemuha

B. 6 atoms of carbon C
I would think this is the answer, because one can't just delete or add atoms; otherwise, the equation would be unbalanced. This also abides with the Law of Conservation of Mass.

Plus, I also came to that conclusion because if we look at the net equation of photosynthesis:
$6C O_{2} + 6H_{2}O$ ––> $C_{6}H_{12}O_{6} + 6O_{2}$
The number of Carbon atoms is 6 on both the reagent's side and the product's side.

8 0

3 years ago

Read 2 more answers

Boyle's Law Problems <br>Charles' Law Problems

zysi [14]

Answer:

here are the answers babe. Feel free to ask for more

8 0

3 years ago

This planet is second from the Sun. It is named after the Roman god of beauty.

lesya [120]

Mercury bc it is melting and getting closer

7 0

4 years ago

Suppose that 98.0g of a non electrolyte is dissolved in 1.00kg of water. The freezing point of this solution is found to be -0.4

Sholpan [36]

Answer:

$\large \boxed{\text{392 u}}$

Explanation:

1. Calculate the molal concentration

The formula for the freezing point depression by a nonelectrolyte is

$\Delta T_{f} = -K_{f}b\\b = -\dfrac{\Delta T_{f}}{ K_{f}} = -\dfrac{-0.465 \, ^{\circ}\text{C}}{\text{1.86 $\, ^{\circ}$C$\cdot$kg$\cdot$mol}^{-1}} = \text{0.250 mol/kg}$

2. Calculate the moles of solute

$\begin{array}{rcl}b & = & \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}\\\\\text{moles of solute} & = & b \times {\text{kilograms of solvent}}\\n & = &\text{0.250 mol/kg} \times \text{1.00 kg}\\ & = & \text{0.250 mol}\\\end{array}$

3. Calculate the molecular mass

$\begin{array}{rcl}\text{Moles} & = &\dfrac{\text{mass}}{\text{molar mass}}\\\\\text{0.250 mol} & = & \dfrac{\text{98.0 g}}{MM}\\\\MM & = & \dfrac{\text{98.0 g }}{\text{0.250 mol}}\\\\& = &\textbf{392 g/mol}\\\end{array}\\\text{The molecular mass of the solute is $\large \boxed{\textbf{392 u}}$}$

3 0

4 years ago