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emmasim [6.3K]
3 years ago
5

A 76 pound flask a mercury cost $126 the density and mercury is 13.534 g/cm³ find the price of one cubic inch of mercury by calc

ulating the following intermediate values
Chemistry
1 answer:
swat323 years ago
3 0
To determine the cost of the mercury per cubic inch, we need to divide the total cost with the total volume in units of cubic inches. To do this, we first determine the volume of the mercury given the mass and the density. In any operation, it is important to remember that the units of the values involved should be homogeneous so that we can cancel them. We do as follows:

mass of mercury = 76 lb ( 1 kg / 2.2 lbs ) ( 1000 g / 1 kg ) = 34545.45 g
volume of mercury in cm^3 = 34545.45 g / 13.534 g / cm^3 = 2552.49 cm^3

We need to convert this to units of cubic inches since it is what is asked.

volume of mercury in in^3 = 2552.49 cm^3 ( 1 in / 2.54 cm )^3 = 155.76 in^3
cost per in^3 = $126 / 155.76 in^3 = $ 0.809 / in^3
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Magnesium burns in air with a dazzling brilliance to produce magnesium oxide: 2 Mg(s) + O2(g) → 2 MgO(s) How many moles of O2 ar
TEA [102]

Answer:

1.05 mols of O2 gas

Explanation:

For this type of problem, it's important to understand what the balanced Chemical equation tells us:

<h3><u>Balanced Chemical equations</u></h3>

Let's look at the balanced chemical equation:

2Mg(s)+O_2(g)-- > 2MgO(s)

This equation two sides of the reaction arrow.  On the left are the reactants (things you start with, and that react during the chemical reaction), and on the right side are products (things that are produced during the chemical reaction that you end with).

The numbers in front of each compound tell how many of molecules are involved for a full reaction without anything left over.  A "mol" is a large quantity (6.022*10^{23}), in this case, of molecules , since it's unlikely you're only taking a single molecule of each substance (it would be so tiny, you wouldn't even know you were doing the reaction).

So,for every 2 moles of Magnesium used, we'll also need 1 mole of Oxygen, and it will produce 2 moles of Magnesium Oxide.

In a way, during the reaction it's almost like 2 moles of Magnesium is equal to 1 mole of Oxygen and is equal to 2 moles of Magnesium Oxide:

2 mol Mg(s)=1mol O_2(g)=2molMgO(s)

From here, we can build some unit ratios, to convert between the known quantity of moles we have, and find the unknown quantity of moles that are requested.

<h3><u>Finding the right unit ratio</u></h3>

We know that we are looking at 2.10 mol of Magnesium, so we want a unit ratio with moles of Mg on the bottom.  <u>We want to find moles of O2</u>, <u>so we want a unit ratio of moles of O2 on top</u>.

The unit ratio we want is the middle part of the equation, divided by the left part of the equation.

2 mol Mg(s)=1mol O_2(g)\\\frac{2 mol Mg(s)}{2 mol Mg(s)}=\frac{1mol O_2(g)}{2 mol Mg(s)}\\1=\frac{1mol O_2(g)}{2 mol Mg(s)}

Since this quantity is 1, it is a unit ratio and can be multiplied to other things to change their units (for this problem).

<h3><u>Finding the answer</u></h3>

Starting with what we know, and multiplying by our unit ratio:

2.10 mol Mg(s)*\frac{1mol O_2(g)}{2 mol Mg(s)}=1.05mol O_2(g)

Notice that the units from the first quantity cancel with the units on the bottom of the fraction, leaving only the unit on top of the fraction ... the exact units we wanted!

So, 1.05mols of O2 would be consumed during the reaction if exactly 2.10moles of Magnesium are burned.

4 0
2 years ago
A 1.36 kg hammer at 22 ms what was the force generated by the hammer
podryga [215]
From the information given, the mass of the hammer seems to be 1.36 kg and the velocity by which it is wielded seems to be 22 m/s. The most suitable formula to be used here is : f = m × v where f is force, m is mass and v is velocity. f = m × v → f = 1.36 × 22 → f = 29.92 kg m/s Therefore the force generated by the hammer is 29.92 N ( kg m/s is the same as Newton)
6 0
3 years ago
A sample of O2(g) is placed in an otherwise empty, rigid container at 4224 K at an initial pressure of 4.97 atm, where it decomp
frutty [35]

Answer:

The value of K_p at 4224 K is 314.23.

Explanation:

O_2(g)\rightleftharpoons 2O(g)

Initially

4.97 atm            0

At equilibrium

4.97 - p               2p

At initial stage, the partial pressure of oxygen gas = =4.97 atm

At equilibrium, the partial pressure of oxygen gas = p_{O_2}=0.28 atm

So, 4.97 - p = 0.28 atm

p = 4.69 atm

At equilibrium, the partial pressure of O gas = p_{O}=2p=2\times 4.69 atm=9.38 atm

The expression of K_p is given as :

K_p=\frac{(p_{O})^2}{(p_{O_2})}

K_p=\frac{(9.38 atm)^2}{0.28 atm}=314.23

The value of K_p at 4224 K is 314.23.

5 0
3 years ago
A boy found a solid metal box in his backyard. The box had been buried for so long, it was difficult to determine from what the
Alex777 [14]

Answer:

Box is made up of <em>copper</em>, because density is <em>8.96  g/cm³.</em>

Explanation:

Given data:

Volume of box = 17.63 cm³

Mass of box = 158 g

Which metal box is this = ?

Solution:

First we will calculate the density of box then we will compare it with the density value of given metals.

d = m/v

d = 158 g/ 17.63 cm³

d = 8.96  g/cm³

The calculated density is similar to the given density value of copper thus box is made up of copper.

4 0
4 years ago
Help please ASAP I’ll mark you as brainlister
noname [10]

Answer: FeO is called ferrous oxide while Fe2O3 is ferric oxide

Explanation:

Ferrous oxide, commonly known as iron(II) oxide contains iron that lost 2 elections in the oxidation process. So it is able to bond with other atoms that have an extra 2 electrons to share. Ferric oxide, is commonly known as iron(III) oxide

8 0
3 years ago
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