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Ratling [72]
3 years ago
8

Atoms of noble gases are generally inert because..

Chemistry
1 answer:
umka21 [38]3 years ago
4 0

Answer:

B. their outer electron levels are filled

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25 POINTS! Which traits are characteristic of both single-celled and multicellular organisms? Choose all answers that are correc
worty [1.4K]
The answer is d. The answer is that because both of them can reproduce that way.
3 0
4 years ago
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What is the percentage composition of each element in hydrogen peroxide, H2O2?
Degger [83]

Answer:

The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).

Explanation:

Step 1: Data given

Molar mass of H = 1.0 g/mol

Molar mass of O = 16 g/mol

Molar mass of H2O2 = 2*1.0 + 2*16 = 34.0 g/mol

Step 2: Calculate % hydrogen

% Hydrogen = ((2*1.0) / 34.0) * 100 %

% hydrogen = 5.88 %

Step 3: Calculate % oxygen

% Oxygen = ((2*16)/34)

% oxygen = 94.12 %

We can control this by the following equation

100 % - 5.88 % = 94.12 %

The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).

8 0
3 years ago
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If a radioactive material has a 10 year half-life, how much of a 100 g sample will be left after 30 years?
olga_2 [115]
12.5g, each 10 years you lose a half of what you have at that given moment

5 0
3 years ago
Which of the following is evidence that groundwater cause erosion and deposition
defon

Answer:You didn't provide examples

Explanation: Groundwater can cause erosion under the surface as it moves through the soil. During the movement an acid is formed which what causes erosion and deposition.

Hope that helps

4 0
3 years ago
A 20.0 g piece of a metal is heated and place into a calorimeter containing 250.0 g of water initially at 25.0 oC. The final tem
BartSMP [9]

Answer:

Q_{metal} = -6799\,J

Explanation:

By the First Law of Thermodynamics, the piece of metal and water reaches thermal equilibrium when water receives heat from the piece of metal. Then:

Q_{metal} = - Q_{w}

Q_{metal} = m_{w} \cdot c_{p,w}\cdot (T_{1}-T_{2})

Q_{metal} = (250\,g)\cdot \left(4.184\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (25\,^{\textdegree}C - 31.5\,^{\textdegree}C)

Q_{metal} = -6799\,J

6 0
3 years ago
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