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liraira [26]
3 years ago
12

Write down the common name of the elements shown in the given table . IfFclBrI​

Chemistry
2 answers:
ipn [44]3 years ago
8 0

Answer:

if you just mean the names here u go

Explanation:

fluorine

chlorine

bromine

iodine

Nat2105 [25]3 years ago
5 0

Answer:

f is the corrrect answer

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How many grams of h3po4 are in 255 ml of a 4.50 m solution of h3po4?
murzikaleks [220]

H3PO4 has molecular weight of approximately 98 grams per mole. 4.50 M is equal to 4.50 mole per 1000 mL solution of H3PO4. 255 mL times 4.50 mol /1000 mL times 98 g/mol is equal to 112.455 grams. Note that I automatically equate 1 Liter to 1000 mL since the given volume is in mL for easier computation.

7 0
3 years ago
A. 1720 kJ<br> B. 125.6 kJ<br> C. 3440 kJ<br> D. 4730 kJ
Feliz [49]

Answer:

Q = 3440Kj

Explanation:

Given data:

Mass of gold = 2kg

Latent heat of vaporization = 1720 Kj/Kg

Energy required to vaporize 2kg gold = ?

Solution:

Equation

Q= mLvap

It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg

by putting values,

Q= 2kg ×  1720 Kj/Kg

Q = 3440Kj

7 0
3 years ago
Identify the group of elements that corresponds to each of the following generalized electron configurations and indicate the nu
Alja [10]

Answer:

(a) [Noble gas] ns² np⁵: Number of unpaired electron is 1 and belongs to group 17 i.e. halogen group of the periodic table.

(b) [noble gas] ns² (n-1)d²: Number of unpaired electron is 2 and belongs to group 4 of the periodic table.

(c) [noble gas] ns² (n-1)d¹⁰ np¹: Number of unpaired electron is 1 and belongs to group 13 of the periodic table.

(d)[noble gas] ns² (n-2)f⁶ : Number of unpaired electron is 6 and belongs to group 8 of the periodic table.

Explanation:

In the Periodic table, the chemical elements are arranged in 7 rows, called periods and 18 columns, called groups. They are organized in increasing order of atomic numbers.

(a) [Noble gas] ns² np⁵

As the total number of electrons in the p-orbital is 5. Therefore, the number of unpaired electron is 1.

This element has 2 electrons in ns orbital and 5 electrons in np orbital. <u>So there are 7 valence electrons.</u>

Therefore, this element belongs to the group 17 i.e. halogen group of the periodic table.

(b) [noble gas] ns² (n-1)d²

As the total number of electrons in the d-orbital is 2. Therefore, the number of unpaired electrons is 2.

This element has 2 electrons in ns orbital and 2 electrons in (n-1)d orbital. So there are <u>4 valence electrons.</u>

Therefore, this element belongs to the group 4 of the periodic table.

(c) [noble gas] ns² (n-1)d¹⁰ np¹

As the total number of electrons in the p-orbital is 1. Therefore, the number of unpaired electron is 1.

This element has 2 electrons in ns orbital and 1 electron in np orbital. So there are <u>3 valence electrons</u>.

Therefore, this element belongs to the group 13 of the periodic table.

(d)[noble gas] ns² (n-2)f⁶

As the total number of electrons in the f-orbital is 6. Therefore, the number of unpaired electron is 6.

This element has 2 electrons in ns orbital and 6 electrons in (n-2)f orbital. So there are <u>8 valence electrons.</u>

Therefore, this element belongs to the group 8 of the periodic table.

3 0
3 years ago
Is a book sitting high upon a shelf a balanced force or an unbalanced force
Art [367]
Balanced force I think
7 0
3 years ago
Read 2 more answers
Suppose a grill lighter contains 50.0 g of butane. How many grams of butane in the lighter would have to be burned to produce 17
Hitman42 [59]

<span>Answer is: mass of burned butane is 11.6 g.</span>

Chemical reaction: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O.

m(butane) = 50,0 g.

<span> V(CO</span>₂) = 17,9 L.<span>
n(CO</span>₂) = V(CO₂) ÷ Vm.<span>
n(CO</span>₂) = 17,9 L ÷ 22,4 L/mol.<span>
n(CO</span>₂) = 0,8 mol.<span>
From chemical reaction n(CO</span>₂) : n(C₄H₁₀) = 8 : 2.<span>
n(C</span>₄H₁₀) = 0,8 mol ÷ 4.<span>
n(C</span>₄H₁₀) = 0,2 mol.<span>
m(C</span>₄H₁₀) = n(C₄H₁₀) · M(C₄H₁₀).<span>
m(C</span>₄H₁₀) = 0,2 mol · 58 g/mol.<span>
m(C</span>₄H₁₀) = 11,6 g.

5 0
3 years ago
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