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2 years ago

What is the mathematical equation used when solving calorimetry problems? What does each variable in the equation represent?

Chemistry

1 answer:

liubo4ka [24]2 years ago

7 0

Answer : The mathematical equation used when solving calorimetry problems is:

$q=m\times c\times \Delta T}$

Explanation :

Calorimetry : It is determining the changes in the energy of a system by measuring the heat transferred with surroundings.

Formula used :

$q=m\times c\times \Delta T}$

where,

c = specific heat capacity of calorimeter

m = mass of a substance

q = heat required

$\Delta T$ = change in temperature of substance

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The hydrogen gas formed in a chemical reaction is collected over water at 30.0 ∘c at a total pressure of 734 mmhg . part a what

In-s [12.5K]

At a temperature of 30 deg C, the vapour pressure of water H2O is about 32 mm Hg. Therefore at a total pressure f 734 mm Hg, the partial pressure of the Hydrogen gas collected is:

<span>P Hydrogen = 734 mm Hg – 32 mm Hg = 702 mm Hg</span>

6 0

3 years ago

Read 2 more answers

Using the mmoles of (35)-2,2,-dibromo-3,4-dimethylpentane calculated earlier and the molecular weight of the product (962 g/mol)

Sedbober [7]

Answer:

The yield of the product in gram is $\mathsf{{w_P}=0.26 \ gram}$

Explanation:

Given that:

the molecular mass weight of the product = 96.2 g/mol

the molecular mass of the reagent (3S)-2,2,-dibromo-3,4-dimethylpentane is 257.997 g

given that the millimoles of the reagent = 2,7 millimoles = $2.7 \times 10^{-3} \ moles$

We know that:

Number of moles = mass/molar mass

Then:

$2.7 \times 10^{-3} = \dfrac{ mass}{257.997}$

$mass = 2.7 \times 10^{-3} \times 257.997$

mass = 0.697

Theoretical yield = (number of moles of the product/ number of moles of reactant) × 100

i.e

Theoretical yield = $\dfrac{n_P}{n_R}\times 100\%$

where;

$n_P = \dfrac{w_P}{m_P}$ and $n_R = \dfrac{w_R}{m_R}$

Theoretical yield = $\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%$

Given that the theoretical yield = 100%

Then:

$100\% =\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%$

$\dfrac{w_P}{m_P}=\dfrac{w_R}{m_R}$

${w_P}=\dfrac{w_R \times m_P}{m_R}$

where,

$w_P$ = derived weight of the product

$m_P =$ the molecular mass of the derived product

$m_R =$ the molecular mass of the reagent

$w_R$ = weight in a gram of the reagent

${w_P}=\dfrac{w_R \times m_P}{m_R}$

${w_P}=\dfrac{0.697 \times 96.2}{257.997}$

$\mathsf{{w_P}=0.26 \ gram}$

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2 years ago

Which kind of statement distinguishes a term from all other terms?

Goshia [24]

I believe the answer is a definition.

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2 years ago

What other signs provide evidence of chemical reactions and show that a chemical change has occurred?

Dmitry_Shevchenko [17]

Color change
new substance formed
heat or light change
new smell
irreversible

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2 years ago

Dogs use the same glycolysis and cellular processes as humans use to produce ATP. A young dog has never had much energy. He is b

scZoUnD [109]

Answer:

<h2>Dog's mitochondria lack the transport protein that transport pyruvate ( end product of glycolysis) across the outer mitochondrial membrane .</h2>

Explanation:

1. As given here that dog's mitochondria can use only fatty acids and also amino acids for their respiration, and as compared to others, Dong's cell produce more lactate then normal, this indicate that his mitochondrial membrane is different then others.

2. The aerobic phases of cellular respiration in eukaryotes occur within mitochondria. These aerobic phases are the TCA Cycle and the electron transport chain. Glycolysis occurs in the cytoplasm and the products of glycolysis enter into the mitochondria to continue cellular respiration.

3. These condition shows that dog's mitochondria lack the transport protein of mitochondria that moves pyruvate across the outer mitochondrial membrane.

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2 years ago