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bogdanovich [222]
2 years ago
14

1. What is the specific heat (C) of an unknown sample that weighs 5.0 grams, absorbds 250.0j of heat and has a temperature

Chemistry
1 answer:
maks197457 [2]2 years ago
7 0
Note: The question is incomplete. The complete question is given below :
Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C)
Explanation:
1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT
Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C
H = 50 × 0.75 × 3 = 112.5 calories
b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g
H = 50 × 45 = 2250 cal
c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories.
Amount of energy left = 5000 - 2362.5 = 2637.5 cal
The remaining energy is used to heat the liquid
H = mcΔT
Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change
2637.5 = 50 × 0.75 x ΔT
ΔT = 2637.5 / ( 50*0.75)
ΔT = 70.3 °C
Final temperature of sample = (70.3 + 27) °C = 97.3 °C
The substance will be in liquid state at a temperature of 97.3 °C

i hope that this eg gonna help u
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Answer:

The constant density decreases

Explanation:

As the temperature of a solvent increases, the solubility of any gas dissolved in that solvent decreases.

For example:

when the temperature of a river, lake or stream is raised high , due to discharge of hot water from some industrial process the solubility of the oxygen in the water is decreased .The fish and the other organisms that live in the water bodies such as rivers, ponds, lakes etc can survive only in the presence of oxygen and decrease in the concentration of the water due to increased temperature can lead to the death of the fish and this may in turn damage the ecosystem.

In the above example, water is considered as the solvent and the oxygen is considered as the solute. When the temperature of the solvent that is water increases, the solubility of the gas that is oxygen in the solvent decreases.

Therefore the answer is decreases

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A standard drink of beer is 12 ounces


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How many elements are found in the following chemical equation? K + Cl → KCl A. 1 B. 2 C. 3 D. 0
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Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts
SOVA2 [1]

Answer:

\boxed{\text{Mg is the limiting reactant}}

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble the data in one place.

           2Mg + O₂ ⟶ 2MgO

n/mol:    2       5

Calculate the moles of MgO we can obtain from each reactant.

From Mg:  

The molar ratio of MgO:Mg is 2:2

\text{Moles of MgO} = \text{2 mol Mg} \times \dfrac{\text{2 mol MgO}}{\text{2 mol Mg}} = \text{2 mol MgO}

From O₂:  

The molar ratio of MgO:O₂ is 2:1.

\text{Moles of MgO} = \text{5 mol O}_{2} \times \dfrac{\text{2 mol MgO}}{\text{1 mol O}_{2}} = \text{10 mol MgO}\\\\\boxed{\textbf{Mg is the limiting reactant}} \text{ because it gives the smaller amount of MgO}

6 0
3 years ago
18-19. If a sample for spectrophotometric analysis is placed in a 10-cm cell, the absorbance will be 10 times greater than the a
miss Akunina [59]

Answer:

As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.

Explanation:

Absorbance of light by a reagent of concentration c, is given as

A = εcl

A = Absorbance

ε = molar absorptivity

c = concentration of reagent.

l = length of light path or length of the solution the light passes through.

So, if all.other factors are held constant, If a sample for spectrophotometric analysis is placed in a 10-cm cell, the absorbance will be 10 times greater than the absorbance in a 1-cm cell.

But the reagent blank solution is called a blank solution because it lacks the given reagent. A blank solution does not contain detectable amounts of the reagent under consideration. That is, the concentration of reagent in the blank solution is 0.

Hence, the Absorbance is subsequently 0. And increasing or decreasing the path length of light will not change anything. As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.

Hope this Helps!!!

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