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3 years ago

What is the metal that is grouped with metalloids

Chemistry

1 answer:

vlabodo [156]3 years ago

4 0

Answer:

The post-transition metals cluster to the lower left of this line. Metalloids: The metalloids are boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te) and polonium (Po). They form the staircase that represents the gradual transition from metals to nonmetals.

Explanation: i googled it

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Of the following reactions occurring at 25ºC, which one involves the greatest increase in entropy?H2(g) + Cl2(g) = 2 HCl(g)H2O(s

SVETLANKA909090 [29]

Answer: [tex]CO_2(s)\rightarrow CO_2(g)[/tex]

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

$\Delta S$ is positive when randomness increases and $\Delta S$ is negative when randomness decreases.

a) $H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$

2 molesof gas are converting to 2 moles of another gas , thus $\Delta S$ is zero.

b) $H_2O(s)\rightarrow H_2O(l)$

1 mole of solid is converting to 1 mole of liquid, the randomness increases and thus $\Delta S$ is positive.

b) $Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)$

2 moles of ions are converting to 1 mole of solid, the randomness decreases and thus $\Delta S$ is negative

d) $CO_2(s)\rightarrow CO_2(g)$

1 mole of solid is converting to 1 mole of gas, the randomness increases drastically and thus $\Delta S$ is highly positive.

3 0

3 years ago

How many moles of water are produced from 0.500 moles of oxygen gas?

solmaris [256]

As far as I know NONE.....
Explanation:
Sodium peroxide can be thermolyzed to give dioxygen gas...
N
a
2
O
2
(
s
)
+
Δ
→
N
a
2
O
(
s
)
+
1
2
O
2
(
g
)
↑
⏐
⏐
⏐

But with water, we simply get an acid base reaction....
N
a
2
O
2
(
s
)
+
2
H
2
O
(
l
)
→
2
N
a
O
H
(
a
q
)
+
H
2
O
2
(
a
q
)
...

6 0

3 years ago

a 1.25g sample of ore containing iron pyrite (FeS2) was pulverized and ignited in air, converting the FeS2 to Fe2O3 and SO2(g).

svp [43]

Answer:

28.9%

Explanation:

Let's consider the following balanced equation.

2 FeS₂ + 11/2 O₂ ⇒ Fe₂O₃ + 4 SO₂

We can establish the following relations:

The molar mass of Fe₂O₃ is 159.6 g/mol
1 mole of Fe₂O₃ is produced per 2 moles of FeS₂
1 mole of Fe is in 1 mole of FeS₂
The molar mass of Fe is 55.84 g/mol

The amount of Fe in the sample that produced 0.516 g of Fe₂O₃ is:

$0.516gFe_{2}O_{3}.\frac{1molFe_{2}O_{3}}{159.6gFe_{2}O_{3}} .\frac{2molFeS_{2}}{1molFe_{2}O_{3}} .\frac{1molFe}{1molFeS_{2}} .\frac{55.84gFe}{1molFe} =0.361gFe$

The percent of Fe in 1.25 g of the ore is:

$\frac{0.361g}{1.25g} .100\%=28.9\%$

4 0

3 years ago

If 5.57 g of Ag2O is sealed in a 75.0-mL tube filled with 760 torr of N2 gas at 28 ∘C, and the tube is heated to 310 ∘C, the Ag2

Snezhnost [94]

Answer: Total pressure = 7293.2 torr or 9.60 atm

Explanation:

Total pressure = partial pressure of nitrogen + partial pressure of oxygen

The partial pressure due to nitrogen is determined using the equation of Gay-Lussac's law: P₁/T₁=P₂/T₂

P₁ = 760 torr = 1atm, T₁ = 28∘C = (273+28)K = 301k, P₂ = ?, T₂ = 310∘C =(310+273)K = 583K

P₂ = P₁ T₂/ T₁

P₂ = 760 * 583 / 301 = 1472.03 torr

The pressure due to Oxygen gas produced is calculated thus:

Balanced equation of the decomposition of Ag₂O at s.t.p. is as follows;

2Ag₂O ----> 4Ag + O₂(g)

2 moles of Ag₂O produces 1 mole of O₂

molar mass of Ag₂O = (2*108 + 16)g = 232g/mol

molar volume of gas at s.t.p. = 22.4L

2*232g i.e. 464g of Ag₂O produces 22.4L of O₂

5.57g of Ag₂O will produce 5.57g*22.4L/464g = 0.269L or 269mL of O₂

Using the General gas equation P₁V₁/T₁=P₂V₂/T₂

P₁ = 1atm = 760 torr, V₁ = 269mL, T₁=273K, P₂ = ?, V₂= 75mL, T₂ = 583K

P₂ = P₁V₁T₂/V₂T₁

P₂ = 760*269*583 / 75*273

P₂ = 5821.17 torr

Total pressure = (1472.03 + 5821.17) torr

Total pressure = 7293.2 torr or 9.60 atm

7 0

3 years ago

What is the maximum number of moles of N-acetyl-p-toluidine can be prepared from 70. milliliters of 0.167 M p-toluidine hydrochl

Kitty [74]

Answer:

$\large \boxed{\text{0.012 mol}}$

Explanation:

We will need a balanced equation with moles, so let's gather all the information in one place.

CH₃C₆H₄NH₂·HCl + (CH₃CO)₂O ⟶ CH₃C₆H₄NHCOCH₃ + junk

V/mL: 70.

c/mol·L⁻¹: 0.167

For simplicity in writing , let's call p-toluidine hydrochloride A and N-acetyl-p-toluidine B.

The equation is then

A + Ac₂O ⟶ B + junk

1. Moles of A

$\text{Moles of A} = \text{70. mL A}\times \dfrac{\text{0.167 mmol A}}{\text{1 mL A}}= \text{12 mmol A}$

2. Moles of B

The molar ratio is 1 mol B:1 mol A

Moles of B = moles of A = 12 mmol = 0.012 mol

$\text{You can prepare $\large \boxed{\textbf{0.012 mol}}$ of N-acetyl-p-toluidine. }$

3 0

3 years ago