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You are starting a new item and after reading the first part you realize you have no idea how to go about answering it. What sho

bulgar [2K]

Answer: 3

Explanation: You may not always get feedback that would be helpful and requesting the answer would be cheating

6 0

3 years ago

suppose you use a pitcher to to estimate 1.25 qt of water ( experimental volume) . You later measured the volume more precisely

saul85 [17]

The percent error in the initial measurement would be 6.47%

<h3>Percent Error</h3>

Percent error in any measurement is mathematically given as:

absolute error/true measurement x 100%

In this case;

True measurement = 1.34 qt

Absolute Error = 1.34 - 1.25

= 0.09 qt

Percent error = 0.09/1.34 x 100%

= 6.47%

More on percent error can be found here: brainly.com/question/3105259?referrer=searchResults

6 0

3 years ago

The mass number of an atom is

Morgarella [4.7K]

Answer:

Cobalt

Explanation:

I think it is cobalt

5 0

2 years ago

Read 2 more answers

Dioxane, C4H8O2, has an enthalpy of fusion of 145.8 J g–1 and its melting point temperature is 11.0 °C. How much heat is require

kumpel [21]

Answer:

20703.6J

Explanation:

Quantity of heat (Q) = mass of dioxane × enthalpy of fusion

Mass of dioxane = 142g

Enthalpy of fusion of dioxane = 145.8J/g

Q = 142g × 145.8J/g = 20703.6J

5 0

4 years ago

Phosphorus pentachloride decomposes according to this equation. PCl5(g) equilibrium reaction arrow PCl3(g) + Cl2(g) An equilibri

melisa1 [442]

Answer:

PCl₅ = 0.03 X 208 = 6.24g

PCl₃ = 0.05 X 137 =6.85 g

Cl₂ = 0.03X71 = 2.13 g

Explanation:

The equilibrium constant will remain the same irrespective of the amount of reactant taken.

Let us calculate the equilibrium constant of the reaction.

Kc= $\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}$

Let us calculate the moles of each present at equilibrium

$moles=\frac{mass}{molarmass}$

molar mass of PCl₅=208

molar mass of PCl₃=137

molar mass of Cl₂=71

moles of PCl₅ = $\frac{mass}{molarmass}=\frac{4.13}{208}=0.02$

moles of PCl₃= $\frac{mass}{molarmass}=\frac{8.87}{137}=0.06$

moles of Cl₂ = $\frac{mass}{molarmass}=\frac{2.90}{71}=0.04$

the volume is 5 L

So concentration will be moles per unit volume

Putting values

Kc = $\frac{\frac{0.06}{5}\frac{0.04}{5}}{\frac{0.02}{5}}=0.024$

Now if the same moles are being transferred in another beaker of volume 2L then there will change in the concentration of each as follow

$PCl_{5}--->PCl_{3}+Cl_{2}$

Initial 0.02 0.06 0.04

Change -x +x +x

Equilibrium 0.02-x 0.06+x 0.04+x

Conc. (0.02-x)/2 (0.06+x)/2 (0.04+x)/2

Putting values

0.024 = $\frac{(0.06+x)(0.04+x)}{(0.02-x)2}$

Solving

$(0.024(2)(0.02-x)=(0.06+x)(0.04+x)$

$0.00096-0.048x=0.0024+x^{2}+0.1x$

$0.148x+x^{2}+0.00144=0$

x = -0.01

so the new moles of

PCl₅ = 0.02 + 0.01 =0.03

PCl₃ = 0.06-0.01 = 0.05

Cl₂ = 0.04-0.01 = 0.03

mass of each will be:

mass= moles X molar mass

PCl₅ = 0.03 X 208 = 6.24g

PCl₃ = 0.05 X 137 =6.85 g

Cl₂ = 0.03X71 = 2.13 g

5 0

4 years ago