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Luba_88 [7]
3 years ago
7

In 1909, Ernest Rutherford performed an experiment to explore the atomic structure. In his experiment, he projected high-speed α

‎ particles onto a thin gold foil. He found that all α‎ particles did not follow the same path. Most of the particles passed through the foil without any scattering, implying that most of the space in an atom is empty. Some particles were scattered at a large angle, and very few of them scattered back in the direction from which they had come. Based on these observations, Rutherford proposed an atomic model, which is known as Rutherford’s atomic model.
O True
O False
Chemistry
1 answer:
xz_007 [3.2K]3 years ago
6 0

Answer:

True

Explanation:

In 1909 Ernest Rutherford proposed to Geiger and Marsden to conduct an experiment in which they would have to launch alpha particles from a radioactive source against a gold foil a few atoms thick.

The aim of the experiment was to corroborate Rutherford's idea that the particles would pass through the metal foil with little deviation.

Rutherford's interpretation of the experimental results gave rise to a new atomic model, and concluded that the mass of the atom was concentrated in a small region of positive charges that impeded the passage of alpha particles. He suggested a new model in which the atom had a nucleus or center in which the mass and the positive charge are concentrated, and that in the extra nuclear zone are the negatively charged electrons.

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The reaction of 5.40 g of carbon with excess O2 yields 13.6 g of CO2. What is the percent yield of this reaction?
vazorg [7]
C + O2= CO2
n  =  \frac{m}{mw}
n =  \frac{5.4}{12}  \\ n  = 0.45 \: mol \: of \: carbon
n =  \frac{13.6}{12 + 16 \times 2} \\ n =  \frac{13.6}{44}  \\ n = 0.31 \: mol \: of \: carbon \: dioxide
CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %
4 0
3 years ago
Many computer chips are manufactured from silicon, which occurs in nature as SiO2. When SiO2 is heated to melting, it reacts wit
riadik2000 [5.3K]

Answer:

A) SiO2 is the limiting reactant

B) Theoretical yield= 72333.3g

C) % yield =91.5%

Explanation:

SiO2(s) + 2C(s) --------------> Si(s) + 2CO(g)

n(SiO2)= 155000/60 = 2583.33 mols

n(C)= 79000/12= 3291.66 mols

a)SiO2 is the limiting reactant

According to the balanced reaction equation,

60g of SiO2 produced 28g of SiO2

155000g of SiO2 will produce 155000×28/60= 72333.3g

Therefore theoretical yield of Si= 72333.3g

% yield= 66200/72333.3×100/1 =91.5%

5 0
3 years ago
When considering metal complexation with EDTA, if you are comparing 2 metals, the metal with a higher ____________ will react wi
DENIUS [597]

If you are comparing 2 metals, the metal with a higher <u>Number of free ions</u> will react with EDTA first

<h3>What is EDTA ?</h3>

EDTA is a type of chemical which binds certain metal ions such as calcium and magnesium. some of the functions of EDTA  includes:

  • Preventing blood clotting of blood samples
  • prevention of the formation of Biofilm by bacterias

The EDTA will readily react with metals which have a hiogher number of free ions that it can bind with.

Hence we can conclude that If you are comparing 2 metals, the metal with a higher <u>Number of free ions</u> will react with EDTA first.

Learn more about EDTA : brainly.com/question/10818175

7 0
2 years ago
A dark brown binary compound contains oxygen and a metal. It is 13.38% oxygen by mass. Heating it moderately drives off some of
Leto [7]

Answer:

a) Mass of O in compound A = 32.72 g

Mass of O in compound B =  21.26 g

Mass of O in compound C = 15.94 g

b) Compound A = MO2

Compound B = M3O4

Compound C = MO

c) M = Pb

Explanation:

Step 1: Data given

A binairy compound contains oxygen (O) and metal (M)

⇒ 13.38 % O

⇒ 100 - 13.38 = 86.62 % M

After heating we get another binairy compound

⇒ 9.334 % O

⇒ 100 - 9.334 = 90.666 % M

After heating we get another binairy compound

⇒ 7.168 % O

⇒ 100 - 7.168 = 92.832 % M

The first compound has an empirical formula of MO2

⇒ 1 mol M for 2 moles O

Step 2: Calculate amount of metal and oxygen in each

compound A:   M  = m1 *0.8662    O = m1 *0.1338

compound B:   M  = m2 *0.90666    O = m2 *0.09334

compound C:   M  = m3 *0.92832    O = m3 *0.07168

Step 3: Calculate mass of oxygen with 1.000 grams of M

Compound A: 1.000g * 0.1338 m1gO  / 0.8662m1gMetal = 0.1545

Compound B: 1.000g * 0.09334 m2gO  / 0.90666m2gMetal = 0.1029

Compound C: 1.000g * 0.07168 m3gO  / 0.92832m3gMetal = 0.07721

Step 4:

1 mol MO2 has 1 mol M and 2 moles O

m1 = (mol O * 16)/0.1338   m1 = 239.2 grams

1 mol M = 0.8632*239.2 = 206.48

0.90666m2 = 206.48  ⇒ m2 = 227.74 g

0.92832m3 = 206.48  ⇒ m3 = 222.42 g

Step 5: Calculate mass of O

Mass of O in compound A = 239.2 - 206.48 = 32.72 g

Mass of O in compound B = 227.74 - 206.48 = 21.26 g

Mass of O in compound C = 222.42- 206.48 = 15.94 g

Step 6: Calculate moles

Moles of O in compound A ≈ 2

⇒ MO2

Moles of O in compound B = 21.26 / 16 ≈ 1.33

⇒ M3O4

Moles of O compound C = 15.94 /16 ≈ 1 moles

⇒ MO

Step 7: Calculate molar mass

The mass of 1 mol metal is 206.48 grams  ⇒ molar mass ≈ 206.48 g/mol

The closest metal to this molar mass is lead (Pb)

6 0
3 years ago
A chemist pours 1 mol of zinc granules into one beaker and 1 mol of zinc chloride powder into another beaker. What do the two sa
irina1246 [14]
The number of Zn particles (atoms of Zn ) in the first sample, and Zn ions are going to be the same, 
Also mass of Zn are going to be the same, because of electrons are too small, and would not have influence on mass.
6 0
3 years ago
Read 2 more answers
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