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goldenfox [79]
3 years ago
5

In which pair is the one on the left larger than the one on the right?

Chemistry
2 answers:
Nuetrik [128]3 years ago
5 0

Answer:

K is larger than K+

Explanation:

Given that the atomic radius is affected by the number of electrons arround the nucleus  (being the numeber of protons constant). The lower the number of electrons, the stonger is the atraction between them and the protons resulting on a smaller atomic radius.

The K+ is a monovalent cation of K, meaning that it has 1 less electron than K and therefore a smaller radius.

GarryVolchara [31]3 years ago
3 0
K because parent atoms are always larger than their cations(positively charged atoms)
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You have a ballon filled with hydrogen gas which keeps it at a constant pressure, regardless the volume. The initial volume of t
abruzzese [7]

Answer:

619°C

Explanation:

Given data:

Initial volume of gas = 736 mL

Initial temperature = 15.0°C

Final volume of gas = 2.28 L

Final temperature = ?

Solution:

Initial volume of gas = 736 mL (736mL× 1L/1000 mL = 0.736 L)

Initial temperature = 15.0°C (15+273 = 288 K)

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = T₁V₂/V₁  

T₂ = 2.28 L × 288 K / 0.736 L

T₂ = 656.6 L.K / 0.736 L

T₂ = 892.2 K

K to °C:

892.2 - 273.15 = 619°C

7 0
3 years ago
Determine the mass of 5.20 moles of c6h12
Gwar [14]
Mass is equal to moles x molar mass, and the molar mass of C6H12 is 84, therefore the mass is 436.8 g, but 437 rounded to correct significant figures
5 0
3 years ago
Can somebody plz help answer these questions plz (grwde7science) only if u know how to do it thx :3
dem82 [27]
1. Crystallization. Magma cools either underground or on the surface and hardens into an igneous rock.
2. weathering and erosion
3. This happens due to geologic uplift and the erosion of the rock and soil above them. At the surface, metamorphic rocks will be exposed to weathering processes and may break down into sediment.
4. Rock Melting.
Metamorphic rocks underground melt to become magma. If you need more help https://www.ck12.org/earth-science/rock-cycle-processes/lesson/Rocks-and-Processes-of-the-Rock-Cycle-HS-ES/
8 0
3 years ago
5.00 moles of a binary, group 2 oxide are found to have a mass of 521 g. Identify the group 2 metal
andrey2020 [161]

5.00 moles of a binary, group 2 oxide are found to have a mass of 521 g. The group 2 metal is Strontium.

According to question 5 moles of binary group 2 metal oxide mass =521  g

So, 1-mole metal oxide mass will be =521 / 5=104.2 g

Now, as metal oxide is group 2 oxide so metal: oxygen =1: 1

So the mass of metal is =104.2-16=88.2 gm (as the atomic mass of oxygen is 16 g

Therefore, the metal is Strontium(Sr) whose atomic mass is 87.62 g which is nearly 88.2 gm

So, the formula of metal oxide is SrO

The group two metal here is Sr which is strontium and lies in the same group as calcium.

To learn more about group 2 metals, visit:

brainly.com/question/18328178

#SPJ9

4 0
2 years ago
For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the
AlladinOne [14]

Answer:

<h3>1. 10 e⁻</h3>

Oxidation numbers

I₂O₅(s): I (5+); O(2-)

CO(g): C(2+); O(2-)

I₂(s): I(0)

CO₂(g): C(4+); O(2-)

<h3>2. 4 e⁻</h3>

Oxidation numbers

Hg²⁺(aq): Hg(2+)

N₂H₄(aq): N(2-); H(1+)

Hg(l): Hg(0)

N₂(g): N(0)

H⁺(aq): H(1+)

<h3>3. 6 e⁻</h3>

Oxidation numbers

H₂S(aq): H(1+); S(2-)

H⁺(aq): H(1+)

NO₃⁻(aq): N(5+); O(2-)

S(s): S(0)

NO(g): N(2+); O(2-)

H₂O(l): H(1+); O(2-)

Explanation:

In order to state the total number of electrons transferred we have to identify both half-reactions for each redox reaction.

1.  I₂O₅(s) + 5 CO(g) → I₂(s) + 5 CO₂(g)

Oxidation: 10 e⁻ + 10 H⁺(aq) + I₂O₅(s) → I₂(s) + 5 H₂O(l)

Reduction: 5 H₂O(l) + 5 CO(g) → 5 CO₂(g) + 10 H⁺(aq) + 10 e⁻

2. 2 Hg²⁺(aq) + N₂H₄(aq) → 2 Hg(l) + N₂(g) + 4 H⁺(aq)

Oxidation: N₂H₄(aq) → N₂(g) + 4 H⁺(aq) + 4 e⁻

Reduction: 2 Hg²⁺(aq) + 4 e⁻ → 2 Hg(l)

3. 3 H₂S(aq) + 2H⁺(aq) + 2 NO₃⁻(aq) → 3 S(s) + 2 NO(g) + 4H₂O(l)

Oxidation: 3 H₂S(aq) → 3 S(s) + 6 H⁺(aq) + 6 e⁻

Reduction: 8 H⁺(aq) + 2 NO₃⁻(aq) + 6 e⁻ → 2 NO(g) + 4 H₂O

5 0
3 years ago
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