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Levart [38]
2 years ago
7

A 11.0 kg iron weightlifting plate has a volume of 1400 cm3 . What is the density of the iron plate in g/cm3?

Chemistry
1 answer:
qwelly [4]2 years ago
5 0
The easy part about this question is that it already gives you the equation to use. If you're trying to find the weight in g/cm^3, all you have to do is divide the weight in grams (g) by the volume in cubic centimeters (cm^3).

The problem already gives you the volume, so you just need to convert kilograms (kg) into grams (g).

Hint: "kilogram" means 1000 grams. There are 1,000 grams in each kilogram.

So find the weight of the object in grams, then divide that number by the volume in cm^3.
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The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4
Evgesh-ka [11]

In NH3 , let oxidation number of N be x

x + (+1)3 = 0

x = -3

In HNO3 , let oxidation number of N be x

1 + x + (-2)3 = 0

x = +5

In NO2 , let oxidation number of N be x

x + (-2)2 = 0

x = +4
5 0
2 years ago
Describe how you could determine the specific heat of a sample of a solid substance. You may assume that the substance does not
lesya692 [45]
<span>I would measure the mass of the solid substance. I would prepare a known mass of room temperature water large enough to submerge the solid substance in question. I would place the water in an insulated container. Then I would heat the solid substance to a known temperature. I would measure the temperature of the heated sample and the water. Then I would submerge the sample in the water and allow the sample and the water to reach the same temperature. I would measure this equilibrium temperature. I would interpret the difference in temperature between the heated sample and the equilibrium temperature as the change in temperature in the sample. Given the known mass, the beginning temperature of the water, and the equilibrium temperature I can determine how much energy was transferred from the heated sample to the water. Now the mass of the sample, a change in temperature in the solid substance, and the amount of energy transferred to create the temperature is known. This is sufficient to determine the specific heat of the solid substance</span>
6 0
2 years ago
Read 2 more answers
Aluminum oxide has a composition of 52.9% aluminum and 47.1% oxygen by mass. if 16.4 g of aluminum reacts with oxygen to form al
Dafna1 [17]
The balanced chemical reaction is written as:

4Al + 3O2 = 2Al2O3

To determine the mass of oxygen gas that would react with the given amount of aluminum metal, we use the initial amount and relate this amount to the ratio of the substances from the chemical reaction. We do as follows:

moles Al = 16.4 g ( 1 mol / 26.98 g ) = 0.61 mol Al
moles O2 = 0.61 mol Al ( 3 mol O2 / 4 mol Al ) = 0.46 mol O2
mass O2 = 0.46 mol O2 ( 32.0 g / mol ) = 14.59 g O2

Therefore, to completely react 16.4 grams of aluminum metal we need a minimum of 14.59 grams of oxygen gas.
3 0
2 years ago
Consider the balanced equation below. 4NH3 + 3O2 --&gt; 2N2 + 6H2O What is the mole ratio of NH3 to N2?
AURORKA [14]
The balanced equation given is:
4NH3 + 3O2 .....> 2N2 + 6H2O

From this equation, we can note that 4 moles of NH3 are required to produce 2 moles of N2.

Therefore, the mole ratio of NH3 to N2 is 4:2 which can be simplified into 2:1
4 0
2 years ago
Read 2 more answers
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N
Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

Now we have to calculate the dissociated concentration of N_2O_4.

The balanced equilibrium reaction is,

                             N_2O_4(g)\rightleftharpoons 2NO_2(aq)

Initial conc.           1.0 M          0

At eqm. conc.     (1.0-x) M    (2x) M

As we are given,

The percent of dissociation of N_2O_4 = \alpha = 28.0 %

So, the dissociate concentration of N_2O_4 = C\alpha=1.0M\times \frac{28.0}{100}=0.28M

The value of x = C\alpha = 0.28 M

Now we have to calculate the concentration of N_2O_4\text{ and }NO_2 at equilibrium.

Concentration of N_2O_4 = 1.0 - x  = 1.0 - 0.28 = 0.72 M

Concentration of NO_2 = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

Now put all the values in this expression, we get :

K_c=\frac{(0.56)^2}{0.72}=0.44

Therefore, the equilibrium constant K_c for the reaction is, 0.44

8 0
2 years ago
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