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Arisa [49]
3 years ago
15

Which groups on the periodic table contain metalloids? Groups 1–2 Groups 3–12 Groups 13–16 Groups 17–18

Chemistry
2 answers:
musickatia [10]3 years ago
7 0

Answer:

13-16

Explanation:

agasfer [191]3 years ago
6 0
The correct answer is groups 13-16
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What is the mass in grams, if you have 8.42 x 10^18 atoms of Bromine
Dahasolnce [82]

Answer:

84.2 grams

Explanation:

you have to multiply 8.42x10^18 to get the answer above

8 0
3 years ago
Nitrogen (N2) and hydrogen (H2) react to form ammonia (NH3). Consider a mixture of six nitrogen molecules and six hydrogen molec
Nezavi [6.7K]

Answer:

a)  No molecules of hydrogen

b) four molecules of ammonia

c) four left molecules of nitrogen.

Explanation:

The balanced reaction between nitrogen and hydrogen molecules to give ammonia molecules is:

N_{2}(g)+3H_{2}(g) -->2NH_{3}

Thus one molecule of nitrogen will react with three molecules of hydrogen to give two molecules of ammonia.

We have six molecules of each nitrogen and hydrogen in the closed container and they undergo complete reaction it means the limiting reagent is hydrogen. For six molecules of nitrogen, eighteen molecules of hydrogen will be required.

So six molecules of hydrogen will react with two molecules of nitrogen to give four molecules of ammonia.

The product mixture will have

a) No molecules of hydrogen

b) four molecules of ammonia

c) four left molecules of nitrogen.

7 0
3 years ago
I need help with these two questions
natulia [17]
I'm not sure about 2 but in 3 it would float
8 0
3 years ago
If we start with 8000 atoms of radium-226 how much would remain after 3200 years?
olya-2409 [2.1K]
<h3>Answer:</h3>

2000 atoms

<h3>Explanation:</h3>

We are given the following;

Initial number of atoms of radium-226 as 8000 atoms

Time taken for the decay 3200 years

We are required to determine the number of atoms that will remain after 3200 years.

We need to know the half life of Radium

  • Half life is the time taken by a radio active material to decay by half of its initial amount.
  • Half life of Radium-226 is 1600 years
  • Therefore, using the formula;

Remaining amount = Original amount × 0.5^n

where n is the number of half lives

n = 3200 years ÷ 1600 years

 = 2

Therefore;

Remaining amount = 8000 atoms × 0.5^2

                                = 8000 × 0.25

                                 = 2000 atoms

Thus, the number of radium-226 that will remain after 3200 years is 2000 atoms.

6 0
3 years ago
A metallurgical firm wishes to dispose of 1200 gallons of waste sulfuric acid whose molarity is 1.05 M. Before disposal, it will
Reil [10]

Answer:

The balanced chemical equation for this process:

H_2SO_4(aq)+Ca(OH)_2(s)\rightarrow 2H_2O(l)+CaSO_4(aq)

$194.51 is the cost that the firm will incur from this use of slaked lime.

Explanation:

The balanced chemical equation for this process

H_2SO_4(aq)+Ca(OH)_2(s)\rightarrow 2H_2O(l)+CaSO_4(aq)

Moles of sulfuric acid = n

Volume of sulfuric acid disposed = V = 1200 gallons = 3.785 × 1200 L = 4,542 L

1 gallon = 3.785 Liter

Morality of the sulfuric acid = M = 1.05 m

Molarity(M)=\frac{n}{V(L)}

n=m\times V=1.05 M\times 4,542 L=4,769.1 mol

According to reaction, 1 mol of sulfuric acid reacts with 1 mole of calcium hydroxide.Then 4,769.1 moles of sulfuric acid will recat with ;

\frac{1}{1}\times 4,769.1 mol=4,769.1 mol of calcium hydroxide

Mass of 4,769.1 moles of calcium hydroxide:

4,769.1 mol  74 g/mol = 352,913.4 g

= \frac{352,913.4 }{453.6} pounds =778.03 pounds

(1 pound = 453.6 grams)

Cost of 1 pound of slaked lime  = $0.25

Cost of 778.03 pounds of slaked lime  = $0.25 × 778.03 = $194.51

$194.51 is the cost that the firm will incur from this use of slaked lime.

3 0
3 years ago
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