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NeX [460]
3 years ago
13

When a student chemist transferred the metal to the calorimeter, some water splashed out of the calorimeter. will this technique

error result in the specific heat of the metal being reported too high or too low?
Chemistry
1 answer:
Natalka [10]3 years ago
5 0

Answer:

It will be reported too low.

Explanation:

To measure the specific heat of the metal (s), the calorimeter may be used. In it, the metal will exchange heat with the water, and they will reach thermal equilibrium. Because it can be considered an isolated system (there're aren't dissipations) the total amount of heat (lost by metal + gained by water) must be 0.

Qmetal + Qwater = 0

Qmetal = -Qwater

The heat is the mass multiplied by the specific heat multiplied by the temperature change. If c is the specific heat of the water:

m_metal*s*ΔT_metal = - m_water *c*ΔT_water

s = -m_water *c*ΔT_water / m_metal*ΔT_metal

So, if m_water is now less than it was supposed to be, s will be reported too low, because they are directly proportional.

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Calculate the molarity of the two solutions. The first solution contains 0.550 mol of NaOH in 1.40 L of solution.
olya-2409 [2.1K]

Answer:

0.393 mol/L.

Explanation:

The following data were obtained from the question:

Number of mole of NaOH = 0.550 mol

Volume of solution = 1.40 L

Molarity of NaOH =.?

Molarity of a solution is simply defined as the mole of solute per unit litre of the solution. Mathematically, it is expressed as:

Molarity = mole /Volume

With the above formula, we can obtain the molarity of the NaOH solution as follow:

Number of mole of NaOH = 0.550 mol

Volume of solution = 1.40 L

Molarity of NaOH =.?

Molarity = mole / Volume

Molarity of NaOH = 0.55 / 1.4

Molarity of NaOH = 0.393 mol/L

Thus, the molarity of the NaOH solution is 0.393 mol/L.

6 0
3 years ago
An enzyme with molecular weight of 310 kDa undergoes a change in shape when the substrate binds. This change can be characterize
oee [108]

Answer:

(a) r = 6.26 * 10⁻⁷cm

(b) r₂ = 6.05 * 10⁻⁷cm

Explanation:

Using the sedimentation coefficient formula;

s =  M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle

s = M ( 1 - Vρ) / N*6πnr

making r sbjct of formula, r =  M (1 - Vρ) / N*6πnrs

Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s

r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)

r = 6.26 * 10⁻⁷cm

b. Using the formula r₂/r₁ = s₁/s₂

s₂ = 0.035 + 1s₁ = 1.035s₁

making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁

r₂ = 6.3 * 10⁻⁷cm / 1.035

r₂ = 6.05 * 10⁻⁷cm

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