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
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
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/
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
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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
