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steposvetlana [31]
3 years ago
8

What is the mass, in grams, of a 12.0cm³ sample of aluminum? The density of aluminum is 2.70g/cm³

Chemistry
1 answer:
Levart [38]3 years ago
4 0

Answer:

The answer is

<h2>32.4 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of aluminum = 12 cm³

Density = 2.70 g/cm³

The mass of aluminum is

mass = 2.7 × 12

We have the final answer as

<h3>32.4 g</h3>

Hope this helps you

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What states can electrons exist in
Norma-Jean [14]

Electrons does not have state like matter of Solid, liquid, gas or plasma states. Electrons together with other molecular and atom components makes up matter. Their characteristic and elemental inclination determines the state for which the matter will exist in.

8 0
3 years ago
Discuss some of the biotic (living) and abiotic (nonliving) factors in the chimps’ ecosystem that affect their behavior.
Genrish500 [490]

Answer:

Diet:  fruit, leaves, bark, stems, seeds, eggs, insects, birds, small to medium sized primates - red tail monkeys, yellow baboons, bushbuck and warthogs.

Environmental Relationship - The chimpanzee keeps the plants it eats short, moves dirt around which helps things living in the dirt, keeps bird and small monkey populations that it eats from overpopulating.

Different biotic and abiotic factors affect why the chimps live where they do. (Spatial Relationships)

Explanation:

8 0
3 years ago
Which balanced equation represents a redox reaction?
nlexa [21]

Answer: option (3) CuO + CO ⇄ Cu + CO₂


Explanation:


1) A redox reaction is an oxidation-reduction reaction.


2) An oxidation-reduction reaction is one in which at least one species is oxidized and other (or others) is reduced.


3) Oxidation is the increase on the oxidation number, which occurs by releasing electrons.


4) Reduction is the decrease of the oxidation number, which occurs by gaining electrons.


5) By looking at the oxidation numbers of the reactants and the products you can tell whether a reaction is a redox one.


6) There are some rules to calculate the oxidation states which you need to handle to determine the oxidation states of the atoms in any chemical formula.


7) The main rule that you can use at a glance is if an element is alone in one side of the equation and is bonded to other kind of atom in the other side.


That is because the rule states that for any element that is not combined with other element its oxidation state is 0.


In this case, you can apply that rule to the equation of the choice number (3). In that, you can see that Cu is pure which as per the mentioned rule means that its oxidation state is 0.


As Cu is combined with O, CuO, in the reactant side of the chemical equation, then its oxidation number is not 0. Indeed, the rule is that in the formula CuO the oxidation states of Cu and O add up 0, and since the oxidation number of O is 2-, the oxidation number of Cu is 2+.


Therefore, having Cu changed its oxidation state from 2+ to 0, it was reduced, and this is a redox reaction.


Of course other atoms had to be oxidized. It was C whose oxidation state went from 2+ in CO to 4+ in CO₂.

7 0
2 years ago
Read 2 more answers
A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is
Colt1911 [192]

Answer:

M_{Na^+}=1.36M

M_{Br^-}=1.58M

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+

Once we've got the moles we compute the final volume via:

V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L

Thus, the molarity of the sodium atoms turn out into:

M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

Now, we perform the same procedure but now for the bromide ions:

n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-

Finally, its molarity results:

M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

Best regards.

7 0
3 years ago
What is the Bronsted acid of H2PO4- + OH-. ---- HPO42- + H2O?​
RoseWind [281]

Answer:

The bronsted- Lowry acid is H₂PO₄⁻

Explanation:

Bronsted-Lowry acid  donates a proton (H⁺)

H₂PO₄⁻ + OH⁻ → HPO₄²⁻ + H₂O

In the reaction above, H₂PO₄⁻ is donating the proton to OH⁻ resulting in H₂O and the deprotonated species. This makes it a bronsted-Lowry acid.

7 0
3 years ago
Read 2 more answers
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