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scoray [572]
3 years ago
9

Why was Rutherford's Gold Foil Experiment so important?

Chemistry
1 answer:
Oksanka [162]3 years ago
4 0

Answer:

Answer is 'B' I think

Explanation:

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. Through previous experiments of shooting alpha particles, Rutherford knew they had considerable mass and speed.

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In which pair would both compounds have the same empirical formula? A. H2O and H2O2 B. BaSO4 and BaSO3 C. FeO and Fe2O3 D. C6H12
Kazeer [188]
The pair of both compounds that have the same empirical formula are C6H12O6 and HC2H3O2. The answer is letter D. <span>H2O and H2O2, BaSO4 and BaSO3 and FeO and Fe2O3 do not have the same empirical formula.</span>
4 0
3 years ago
What is the shape of graphite​
VashaNatasha [74]

Answer:

The molecular formula lists the symbol of each element within the compound followed by a number (usually in subscript). The letter and number indicate how many of each type of element are in the compound. If there is only one atom of a particular element, then no number is written after the element.

7 0
3 years ago
4. Does an atom of an element always have the same number of neutrons as other atoms of the same
Sergeeva-Olga [200]

Explanation:

No. Isotopes are atoms of the same element with different atomic masses (due to the different number of neutrons)

For example, carbon exists as carbon-12 and carbon-14, which both have 6 protons but have 6 and 8 neutrons respectively.

6 0
3 years ago
Assuming a car (with a 70-L) gas tank can hold approximately 50,000 (5.00 * 10^4) g of octane(C8H18) or 50,000 (5.00 * 10^4) g o
konstantin123 [22]

Answer:

- From octane: m_{CO_2}=1.54x10^5gCO_2

- From ethanol: m_{CO_2}=9.57x10^4gCO_2

Explanation:

Hello,

At first, for the combustion of octane, the following chemical reaction is carried out:

C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O

Thus, the produced mass of carbon dioxide is:

m_{CO_2}=5.00x10^4gC_8H_{18}*\frac{1molC_8H_{18}}{114gC_8H_{18}}*\frac{8molCO_2}{1molC_8H_{18}}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=1.54x10^5gCO_2

Now, for ethanol:

C_2H_6O+3O_2\rightarrow 2CO_2+3H_2O

m_{CO_2}=5.00x10^4gC_2H_6O*\frac{1molC_2H_6O}{46gC_2H_6O}*\frac{2molCO_2}{1molC_2H_6O}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=9.57x10^4gCO_2

Best regards.

3 0
3 years ago
I really need help with this problem.
Aleksandr [31]

Answer:

.371 mole of NaCl

Explanation:

Na Cl Mole weight = 22.989   + 35.45 = 58.439 g/mole

21.7 g / 58.439 g/mole = .371 mole

8 0
2 years ago
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