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3 years ago

What does a bomb calorimeter do?

2 answers:

7 0

It is a constant-volume type calorimeter that measures the heat of a particular reaction or measures the calorific value of the fuels. Bomb calorimeters are built in such a way that they can withstand the large pressure produced within the calorimeter due to the reaction or burning of fuel.

koban [17]3 years ago

3 0

You might be interested in

If the compound contains 35.06 % cl by mass, what is the identity of the metal?

Setler [38]

Answer is: identity of the metal is gold (Au).

ω(Cl) = 35.06% ÷ 100%.

ω(Cl) = 0.3506; mass percentage of chlorine.

If we take 100 grams of the compound:

m(Cl) = ω(Cl) · m(compound).

ω(Cl) = 0.3506 · 100 g.

ω(Cl) = 35.06 g.

n(Cl) = m(Cl) ÷ M(Cl).

n(Cl) = 35.06 g ÷ 35.45 g/mol.

n(Cl) = 0.99 mol; amount of substance.

In molecule MCl₃: n(M) : n(Cl) = 1 : 3.

n(M) = 0.33 mol; amount of unknown metal.

M(M) = m(M) ÷ n(M).

M(M) = (100 g - 35.06 g) ÷ 0.33 mol.

M(M) = 196.8 g/mol; molar mass of the gold.

4 0

3 years ago

Find the boiling point? 100. g of C2H6O2 dissolved in 200 g of H2O?

aleksklad [387]

Answer:

The correct answer is 104.13ºC

Explanation:

When a solute is added to a solvent, the boiling point of the solvent (Tb) increases. That is a colligative property. The increment in Tb (ΔTb) is given by the following expression:

ΔTb = Tb - Tbº= Kb x m

Where Tb and Tbº are the boiling points of the solvent in solution and pure, respectively; Kb is a constant and m is the molality of the solution.

In this problem, the solvent is water (H₂O). It is well known that water has a boiling point of 100ºC (Tb). The value of Kb for water is 0.512ºC/m. So, we have to calculate the molality of the solution (m):

m = moles of solute/Kg solvent

The solute is C₂H₆O₂ and we have to calculate the number of moles of this component by dividing the mass into the molecular weight (Mw):

Mw(C₂H₆O₂)= (2 x 12 g/mol) + (6 x 1 g/mol) + (2 x 16 g/mol)= 62 g/mol

⇒ moles of C₂H₆O₂ = mass/Mw = 100 g/(62 g/mol) = 1.613 moles

Now, we need the mass of solvent (H₂O) in kilograms, so we divide the grams into 1000:

200 g x 1 kg/1000 g = 0.2 kg

Finally, we calculate the molality as follows:

m = 1.613 moles of C₂H₆O₂/0.2 kg = 8.06 moles/kg = 8.06 m

The increment in the boiling point will be:

ΔTb = Kb x m = 0.512ºC/m x 8.06 m = 4.13ºC

So, the boiling point of pure water (Tbº=100ºC) will increase in 4.13ºC:

Tb= 100ºC+4.12ºC= 104.13ºC

5 0

3 years ago

Two students in the same class find that a piece of wood,

Naily [24]

Answer:

The wood is oak.

Explanation:

Given data:

Mass of wood = 70 g

Volume of wood = 103 cm³

Wood is oak = ?

Solution:

We will calculate the density of wood then we will compare it with literature value.

Density = mass/ volume

d = 70 g/ 103 cm³

d = 0.68 g/cm³

The density of oak is 0.59 - 0.90 g/cm³.

So its true, wood is oak.

5 0

3 years ago

Which actions are part of a system of methodical tests and refinements specific to technological design? 1 documenting, tinkerin

telo118 [61]

Answer:

1. documenting, tinkering, testing

Explanation:

Technological design is defined as the process of study, design and development of new technologies.

There are some action in the methodical tests and refinements specific to technological design include documenting, tinkering, testing.

Documenting includes collecting all the information about the design and develop the product, tinkering involves repairing or adjust the issues found in the development, and testing helps to evaluate if the product is ready to work as it is supposed to.

Hence, the correct answer is "1."

6 0

3 years ago

Consider the intermediate chemical reactions. 2 equations. First: upper C a (s) plus upper C upper O subscript 2 (g) plus one ha

DochEvi [55]

Answer: When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The overall chemical reaction follows:

$CaO(s)+CO_2\rightarrow CaCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\rightarrow CaCO_3(s)$ $\Delta H_1=-812.8kJ$

(2) $2Ca(s)+O_2(g)\rightarrow 2CaO(s)$ $\Delta H_2=-1269kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[\frac{1}{2}\times (-\Delta H_2)]$

Hence, when the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

3 0

3 years ago