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zlopas [31]
3 years ago
13

The day to day condition of atmosphere is known as

Chemistry
2 answers:
vovikov84 [41]3 years ago
5 0

Answer:

d) atmosphere pressure

Sphinxa [80]3 years ago
3 0

Answer:

B.) weather

Explanation:

weather is constantly changing everyday and each day presents a new condition:)

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1. Rank the following compounds in order of decreasing acid strength using periodic trends. Rank the acids from strongest to wea
Softa [21]

Answer:

1. HBr>HCl> H2S >BH3

2.K_a1 very large — H2SO4

K_a1= 1.7 x 10^−2 — H2SO3

K_a1 = 1.7 x 10^−7 — H2S

Explanation:

As one goes down a row in the Periodic Table the properties that determine the acid strength can be observed.

The atoms get larger in radius meaning that in strength, the strength of the bonds get weaker, conversely meaning that the acids get stronger.

For the halogen-containing acids above following the rows and periods, HBr has the strongest bond and is the strongest acid and others follow in this order.

HBr>HCl> H2S >BH3

Acid Dissociation Constant provides us with information known as the ionization constant which comes in handy to measure the acid's strength. The meaning of the proportions are thus, the higher the Ka value, the stronger the acid i.e. it liberates more number of hydrogen ions per mole of acid in solution.

In solution strong acids completely dissociate hence, the value of dissociation constant of strong acids is very high.

Following the cues above on Ka;

K_a1 very large — H2SO4

K_a1= 1.7 x 10^−2 — H2SO3

K_a1 = 1.7 x 10^−7 — H2S

5 0
3 years ago
Assume that the reaction of aqueous hydrobromic acid solution and potassium hydroxide base undergoes a complete neutralization r
Tamiku [17]

Answer:

a.

HBr + KOH → KBr + H  _{2} O

b.

1000 \: ml \: contains \: 0.685 \: moles \\ 55.4 \: ml \: contains \: ( \frac{55.4 \times 0.685}{1000} ) \\  = 0.038 \:moles \\ 1 \: mole \: of \: hydrobromic \: acid  \: produces \: 1 \: mole \: of \: water \\ 0.038 \: moles \: produce \: (0.038 \times 1) \\  = 0.038 \: moles \\ 1 \: mole \: of \: water \: weighs \: 18 \: g \\ 0.038 \: moles \: weighs \: (0.038 \times 18) \: g \\  = 0.684 \: g

c.

0.042 M

d.

1.4

5 0
3 years ago
A certain liquid X has a normal boiling point of 111.20 celsius and a boiling point elevation constant Kb=0.95. calculate the bo
Goshia [24]

Answer: the boiling point is = 137.325°C

Explanation:

From the formula: ∆Tb= Kb*m

From the question, Kb= 0.95, m= 27.5, T1= 111.2°C

Substitute into ∆Tb= Kb*m

∆Tb= 0.95*27.5= 26.125

∆Tb= T2-T1

Hence

T2- 111.2=26.125

T2= 26.125+ 111.2= 137.325°C

8 0
3 years ago
The central Xe atom in the XeF4 molecule has ________ unbonded electron pair(s) and ________ bonded electron pair(s) in its vale
Dimas [21]

Answer:

Lewis structure in attachment.

Explanation:

Atoms of elements in and beyond the third period of the  periodic table form some compounds in which more than eight electrons surround the  central atom. In addition to the 3s and 3p orbitals, elements in the third period also  have 3d orbitals that can be used in bonding. These orbitals enable an atom to form  an <u>expanded octet</u>.

The central Xe atom in the XeF₄ molecule has <u>two</u> unbonded electron pairs and <u>four</u> bonded electron pairs in its valence shell.

8 0
3 years ago
Given the equation I = Q/t, solve for t.
larisa86 [58]

Answer: t=\frac{Q}{I}

Explanation:

I=\frac{Q}{t}

Multiply by t on both sides.

t*I=\frac{Q}{t}*t

tI=Q

Now divide by I to isolate t.

\frac{tI}{I}=\frac{Q}{I}

t=\frac{Q}{I}

5 0
3 years ago
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