During trial 1, the limiting reactant is __________ because____________________________

How does water heat earth

vovikov84 [41]

The answer is the 3rd one down I think

6 0

3 years ago

The plunger on a bicycle pump with a 400 mL volume cylinder is

Kobotan [32]

Answer:

The answer to your question is V2 = 66.7 ml

Explanation:

Data

Volume 1 = V1 = 400 ml

Pressure 1 = P1 = 1 atm

Volume 2 = V2 = ?

Pressure 2 = P2 = 6 atm

Process

1.- To solve this problem use Boyle's law

P1V1 = P2V2

-solve for V2

V2 = P1V1 / P2

-Substitution

V2 = (1)(400) / 6

-Simplification

V2 = 400 / 6

-Result

V2 = 66.7 ml

4 0

3 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers

Suppose that coal of density 1.5 g/cm^3 is pure carbon. (It is, in fact, much more complicated, but this is a reasonable first a

NISA [10]

Answer:

q = -6464.9 kJ

Explanation:

We are given that the heat of combustion is ∆H° = −394 kJ per mol of carbon.Therefore what we need to do is calculate how many moles of C are in the lump of coal by finding its mass since the density is given.

vol = 5.6 cm x 5.1 cm x 4.6 cm = 131.38 cm³

m = d x v = 1.5 g/cm³ x 131.38 cm³ = 197.06 g

mol C = m/MW = 197.06 g/ 12.01g/mol = 16.41 mol

q = −394 kJ /mol C x 16.41 mol C = -6464.9 kJ

7 0

3 years ago

Which reaction has the highest atom economy for the production of C02?

In-s [12.5K]

The highest atom economy

2CO + O₂ ⇒ 2CO₂

<h3>Further explanation</h3>

Given

The reaction for the production of CO₂

Required

The highest atom economy

Solution

In reactions, there are sometimes unwanted products that can be said to be a by-product or a waste product. Meanwhile, the desired product can be said to be a useful product, which can be shown as the atom economy

of the reaction

the higher the atomic economy value of a reaction, the smaller the waste/ byproducts produced, so that less energy is wasted

The general formula:

Atom economy = (mass of useful product : mass of all reactants/products) x 100

Atom economy = (total formula masses of useful product : total formula masses of all reactants/products) x 100

So a reaction that only produces one product will have the highest atomic value, namely the reaction in option C

4 0

3 years ago

During trial 1, the limiting reactant is __________ because_____________________________.

During trial 1, the limiting reactant is because___________________.