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almond37 [142]
3 years ago
5

A 28.5 gram piece of iron is added to a graduated cylinder containing 45.5 mL of water. The water in the cylinder rises to the 4

9.1 mark. Calculate the density of the iron piece.
Chemistry
1 answer:
yan [13]3 years ago
4 0

7.91 g/ml is the density of the iron piece of 28.5 gms.

Explanation:

The density of a substance is defined as the volume it occupies.  It tells the matter present in a substance.

The density is mass per unit volume and is denoted by p.

The formula for density is given by:

density (p) = \frac{mass}{volume}

Data given is :

mass= 28.5 grams

V1 = 45.5 ml

V2= 49.1 ml

The initial volume of water was 45.5 ml, when iron piece of 28.5 grams was added the final volume was 49.1 ml.

Putting the values in the equation of density

p = \frac{28.5}{49.1-45.5}

p = 7.91 g/ml

Since iron is a dense material it will occupy less volume

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In a study of the following reaction at 1200 K it was observed that when the equilibrium partial pressure of water vapor is 15.0
sp2606 [1]

Answer:

The value of K_p for this reaction at 1200 K is 4.066.

Explanation:

Partial pressure of water vapor at equilibrium = p^o_{H_2O}=15.0 Torr

Partial pressure of hydrogen gas at equilibrium = p^o_{H_2}=?

Total pressure of the system at equilibrium P = 36.3 Torr

Applying Dalton's law of partial pressure to determine the partial pressure of hydrogen gas at equilibrium:

P=p^o_{H_2O}+p^o_{H_2}

p^o_{H_2}=P-p^o_{H_2O}=36.3 Torr- 15.0 Torr = 21.3 Torr

3 Fe(s) 4 H_2O(g)\rightleftharpoons Fe_3O_4(s) 4 H_2(g)

The expression of K_p is given by:

K_p=\frac{(p^o_{H_2})^4}{(p^o_{H_2O})^4}

K_p=\frac{(21.3 Torr)^4}{(15.0 Torr)^4}=4.066

The value of K_p for this reaction at 1200 K is 4.066.

6 0
3 years ago
Which of the following is not a property of an acid?
SpyIntel [72]

Answer:

1. bitter  

2. turns litmus paper red  

3. oxygen, hydrogen, and molecules.

Explanation:

7 0
3 years ago
Read 2 more answers
In a titration, a few drops of an indicator are added to a flask containing 35.0 milliliters of HNO3(aq) of unknown concentratio
sasho [114]

Answer:

NaOH(aq) + HNO3(aq)------>NaNO3(aq) + H2O(l)

Explanation:

A thing to note is that an acid and a base will react to form a metal salt + H2O.

~Hope it helps:).

5 0
2 years ago
stbank, Question 075 Get help answering Molecular Drawing questions. Compound A, C6H12 reacts with HBr/ROOR to give compound B,
Law Incorporation [45]

Answer:

Explanation:

In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).

A and C reacts with two differents reagents and conditions, however both of them gives the same product.

Let's analyze each reaction.

First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.

Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.

8 0
3 years ago
Help ASAP pleaseeee thank you
Artyom0805 [142]
The correct answer is B. balance
8 0
2 years ago
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