Answer:
see attached
Explanation:
Dimensional analysis is useful whenever dimensions are involved. Unless it is quite clear that all of the problem dimensions are consistent (for example, all speeds in miles per hour, or all angles in degrees), dimensional analysis can be useful for keeping the math straight.
Only units of the same dimensions can be added or subtracted. When numbers are multiplied or divided or raised to a power, dimensional analysis can help ensure that the appropriate operations are being used on appropriate numbers. It can also help ensure that dimensions are being combined properly to give appropriate derived dimensions.
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Scientific notation is a way of writing very large or very small numbers compactly. It can also help with "order of magnitude" estimates. If an answer using SI prefixes is appropriate, or if a number can be conveniently expressed in standard form, then scientific notation is usually not required.
On the other hand, SI prefixes may not be appropriate in some cases, or a problem may specify that scientific notation be used for expressing results. In those instances, scientific notation should be used.
When the amount of heat gained = the amount of heat loss
so, M*C*ΔTloses = M*C* ΔT gained
when here the water is gained heat as the Ti = 25°C and Tf= 28°C so it gains more heat.
∴( M * C * ΔT )W = (M*C*ΔT) Al
when Mw is the mass of water = 100 g
and C the specific heat capacity of water = 4.18
and ΔT the change in temperature for water= 28-25 = 3 ° C
and ΔT the change in temperature for Al = 100-28= 72°C
and M Al is the mass of Al block
C is the specific heat capacity of the block = 0.9
so by substitution:
100 g * 4.18*3 = M Al * 0.9*72
∴ the mass of Al block is = 100 g *4.18 / 0.9*72
= 19.35 g
Answer:
Empirical formula is Li₂CO₃.
Explanation:
Percentage of oxygen= 65.0%
Percentage of lithium = 18.7%
Percentage of carbon= 16.3%
Empirical formula = ?
Solution:
Number of gram atoms of C = 16.3/12 = 1.4
Number of gram atoms of Li = 18.7/6.94 = 2.7
Number of gram atoms of O = 65.0/ 16 = 4.1
Atomic ratio:
Li : C : O
2.7/1.4 : 1.4/1.4 : 4.1/1.4
2 : 1 : 3
Li : C : O = 2 : 1 : 3
Empirical formula is Li₂CO₃.
Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
Answer:
See explanation below for answers
Explanation:
We know that the balance is tared, so the innitial weight would be zero. Now, let's answer this by parts.
a) mass of displaced water.
In this case all we need to do is to substract the 0.70 with the 0.13 g. so:
mW = 0.70 - 0.13
mW = 0.57 g of water
b) Volume of water.
In this case, we have the density of water, so we use the formula for density and solve for volume:
d = m/V
V = m/d
Replacing:
Vw = 0.57/0.9982
Vw = 0.5710 mL of water
c) volume of the metal weight
In this case the volume would be the volume displaced of water, which would be 0.5710 mL
d) the mass of the metal weight.
In this case, it would be the mass when the metal weight hits the bottom which is 0.70 g
e) density.
using the above formula of density we calculate the density of the metal
d = 0.70 / 0.5710
d = 1.2259 g/mL
