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ki77a [65]
3 years ago
14

Identify the group of ONLY metalloids.

Chemistry
1 answer:
Alinara [238K]3 years ago
5 0

Answer:

C boron, germanium, arsenic, silicon

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Passing an electric current through a sample of water (H2O) can cause the water to decompose into hydrogen gas (H2) and oxygen g
Setler [38]
The balanced chemical reaction is:

<span>2H2O= 2H2 + O2
</span>
We are given the amount of oxygen to be produced in the reaction. The starting point for the calculations will be this amount.

50 g ( 1 mol O2 / 32 g O2 ) ( 2 mol H2O / 1 mol O2 ) ( 18.01 g H2O / 1 mol H2O) = 56.28 g of H2O is needed.

Therefore, the correct answer is the last option.
8 0
3 years ago
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What are the properties of seismic waves?
natima [27]

Answer: incompressibility, rigidity, and density

Explanation:

6 0
2 years ago
Determine the total time that must elapse until only ¼ of an original sample of the radioisotope Rn-222 remains unchanged.
Arte-miy333 [17]

Answer:

7.6 days

Explanation:

Radon is a radioactive element and Radon-222 is it's most stable isotope. The half-life of Radon-222 has been found to be approximately 3.8 days.

Let, the initial amount of the Rn-222 = 1 = A

Final amount = \frac{1}{4} = A'

We will use the following relation for calculating time elapsed in the decay

A' = A(\frac{1}{2} )^\frac{t}{t_1_/_2}  }

Thus,

\frac{1}{4} =1(\frac{1}{2} )^\frac{t}{3.8}

We can write is as,

(\frac{1}{2} )^2=(\frac{1}{2} )^\frac{t}{3.8}

Since the base in both sides are equal, powers can also be equal and thus,

2=\frac{t}{3.8}

So, t = 7.6 days

5 0
3 years ago
Determine the number of representative particles in each of the following.
klio [65]

Answer: 1. 1.59 x 10^23 Particles

2. 4.79 x 10^21 particles

3. 2.67 x 10^25 particles

4. 2.12 x 10^23 Particles

Explanation: 1mole of any substance contains 6.02x10^23 Particles

1. 0.264 mol of silver will contain = 0.264 x 6.02x10^23 = 1.59 x 10^23 Particles

2. 7.95 x 10^-3 mol sodium chloride will contain = 7.95 x 10^-3 X 6.02x10^23 = 4.79 x 10^21 particles

3. 44.4 mol carbon dioxide will contain = 44.4 x 6.02x10^23 = 2.67 x 10^25 particles

4. 0.352 mol nitrogen will contain = 0.352 x 6.02x10^23 = 2.12 x 10^23 Particles

4 0
3 years ago
If 57.0 g of b2o3 is added to 44.7 g of cl2 and 68.8 g of c, what is the theoretical yield of boron trichloride?
sdas [7]
(57.0 g B2O3 / (69.6202 g B2O3/mol) x (4mol BCI3 / 2 mol B2O3) = 1.64 mol BC13

(44.7 g C12) / (70.9064 g C12/mol) x (4mol BCI3 / 6mol C12) = 0.42027 mol BC13

(68.8 g C) / (12.01078 G C/mol) x (4mol BCI3 / 3 mol C) = 7.63 mol BCI3

C12 is the limiting reactant.

(0.42027 mol BCI3) X (117 . 170 g BCI3/mol) = 49.2 g BCI3 in theory.
6 0
3 years ago
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