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Darya [45]
3 years ago
11

An aluminum bar was found to have a mass of 27g. Using water displacement, the volume was measured to be 10 ml. What is the dens

ity of the aluminum? Group of answer choices (27 g)/(10 ml) (10 ml )/(2.70 g) (270 g)/(10 ml) (10 ml )/(27 g)
Chemistry
1 answer:
Pavlova-9 [17]3 years ago
3 0

Answer:

2.7 g/mL:)

An aluminum bar was found to have a mass of 27g. Using water displacement, the volume was measured to be 10 ml. What is the density of the aluminum? Group of answer choices (27 g)/(10 ml) (10 ml )/(2.70 g) (270 g)/(10 ml) (10 ml )/(27 g)

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A Solution is a mixture of two or more substances_____
Ipatiy [6.2K]

Answer:

In the same phase

Explanation:

Because when you make a solution you have to keep them in the same phase or you will oof.

4 0
3 years ago
using the equation you wrote determine how many moles of butane c4h10 are needed to react with 5.5 moles of oxygen
GaryK [48]

Answer:

0.846 moles.

Explanation:

  • This is a stichiometric problem.
  • The balanced equation of complete combustion of butane is:

C₄H₁₀ + 6.5 O₂ → 4 CO₂ + 5 H₂O

  • It is clear from the stichiometry of the balanced equation that complete combustion of 1.0 mole of butane needs 6.5 moles of O₂ to produce 4 moles of CO₂ and 5 moles of H₂O.

<u><em>Using cross multiplication:</em></u>

  • 1.0 mole of C₄H₁₀ reacts with → 6.5 moles of O₂
  • ??? moles of C₄H₁₀ are needed to react with → 5.5 moles of O₂
  • The number of moles of C₄H₁₀ that are needed to react with 5.5 moles of O₂ = (1.0 x 5.5 moles of O₂) / (6.5 moles of O₂) = 0.846 moles.
3 0
3 years ago
How many moles of chromium III nitrate are produced When chromium reacts with 0.85 moles of lead for nitrate to produce chromium
Pepsi [2]

0.85 moles formula units of lead nitrate will produce 0.57 moles formula units of chromium (III) nitrate.

<h3>Explanation</h3>

Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as \text{Pb}^{2+} ions. The formula for a nitrate ion is {\text{NO}_3}^{-}. The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each \text{Pb}^{2+} ion in lead nitrate would pair up with two {\text{NO}_3}^{-} ions. The formula for lead nitrate will be \text{Pb}({\text{NO}_3})_2. Each formula unit of lead nitrate will contain one \text{Pb}^{2+} ion and two {\text{NO}_3}^{-} ions.

The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as \text{Cr}^{3+}. Similarly, each \text{Cr}^{3+} will pair up with three {\text{NO}_3}^{-} ions. The formula for chromium (III) nitrate will be \text{Cr}(\text{NO}_3})_3. Each formula unit of chromium (III) nitrate will contain one {\text{NO}_3}^{-} ion and three {\text{NO}_3}^{-} ions.

0.85 moles formula units of lead nitrate will contain 0.85 × 2 = 1.7 moles of {\text{NO}_3}^{-} ions. Those nitrate ions will end up in 1.7 / 3 = 0.57 moles formula units of chromium (III) nitrate. As a result, the reaction will produce 0.57 moles formula units of chromium (III) nitrate.

7 0
3 years ago
Predict the products of La(s) + O2(aq) -&gt;
mina [271]

Answer:

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

Explanation:

<em>Hello </em><em>there!</em>

When you are given such a problem for completing the chemical equations, what you have to understand is that metals are found in groups I, II and III. While Oxygen is a group VI element.

From the above question I have considered that my La(s), solid is either Sodium (Na) - group I, Magnesium - group II and Aluminum - group III.

In a reaction, there is exchange of electrons given by their oxidation numbers (I, II and III - for our metals above)

The chemical equations are thus;

i. Na (s) + O2 (g) => NaO (s)

ii. Mg (s) + O2 (g) => Mg2O (s)

iii. Al (s) + O2 (g) => Al2O3 (s)

Relate this to the problem and it will be;

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

<em>I hope this </em><em>helps </em><em>you</em><em> </em><em>to </em><em>understand</em><em> </em><em>better</em><em>.</em><em> </em><em>Enjoy </em><em>your</em><em> </em><em>studies</em>

3 0
2 years ago
8. How many atoms of cobalt are in a 0.39 mole sample of CoC
Debora [2.8K]

8.98

×

N

A

 

cobalt atoms

Explanation:

N

A

,

Avogadro's number

specifies  

6.0221

×

10

23

individual particles. It is simply another collective number like a dozen, or a score, or a gross.  

N

A

has the property that  

6.0221

×

10

23

individual cobalt atoms has a mass of  

58.93

⋅

g

. How did I know that? Did I have it memorized?

So the quantity is  

≈

 

54

×

10

23

cobalt atoms.

3 0
3 years ago
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