Based on the periodic table, the element mendeleev called eka-manganese is now called technetium. Technetium<span> is a silvery-gray metal that tarnishes slowly in moist air. Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>
Answer:
1.2029 J/g.°C
Explanation:
Given data:
Specific heat capacity of titanium = 0.523 J/g.°C
Specific heat capacity of 2.3 gram of titanium = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
1 g of titanium have 0.523 J/g.°C specific heat capacity
2.3 × 0.523 J/g.°C
1.2029 J/g.°C
Mean: the average. you have to add the values of the numbers and then divide by the amount of numbers there are. a common mistake to avoid is forgetting to divide the numbers at the end or subtracting them instead of adding.
mean: the middle number. you would first need to order the numbers from least to greatest. a common mistake to avoid is finding the middle number before ordering it from least to greatest
these two can also be commonly mistaken for one another because of the similar spelling.
Answer:
The initial volume in mL is 5959.2 mL
Explanation:
As the number of moles of a gas increases, the volume also increases. Hence, number of moles and volumes are directly proportional i.e
n ∝ V
Where n is the number of moles and V is the volume
Then, n = cV
c is the proportionality constant
∴n/V = c
Hence n₁/V₁ = n₂/V₂
Where n₁ is the initial number of moles
V₁ is the initial volume
n₂ is the final number of moles
and V₂ is the final volume.
From the question,
n₁ = 0.693 moles
V₁ = ?
n₂ = 0.928 moles
V₂ = 7.98 L
Putting the values into the equation
n₁/V₁ = n₂/V₂
0.693 / V₁ = 0.928 / 7.98
Cross multiply
∴ 0.928V₁ = 0.693 × 7.98
0.928V₁ = 5.53014
V₁ = 5.53014/0.928
V₁ = 5.9592 L
To convert to mL, multiply by 1000
∴ V₁ = 5.9592 × 1000 mL
V₁ = 5959.2 mL
Hence, the initial volume in mL is 5959.2 mL
