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3 years ago

Which one of these is an element carbon carbon dioxide air water

1 answer:

8 0

Carbon is an element.
Carbon dioxide is a compound
air is a mixture
water can be both a compound/mixture depending on the type

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What information do the coefficients of a balanced equation give about the reactants

Nana76 [90]

The information that the coefficients of a balanced equation give about the reactants are to tell how many moles of reactants are needed and how many moles of product can be produced.

4 0

3 years ago

How many moless is 4.91x10^22 molecules of H3PO4?

Lynna [10]

1 mol = 6.023x10^23 number of molecules (Avogadro's number)

1 : 6.023x10^23
X : 4.91x10^22

(6.023x10^23)X = 4.91x10^22

X = 4.91x10^22/6.023x10^23

X = 0.082 Moles

6 0

3 years ago

Find the mass that corresponds to each number of

jeka94

Answer:159.7 g.

Explanation:

8 0

3 years ago

Read 2 more answers

All of the elements on periodic table are found in nature true or false

Andreas93 [3]

Answer:

The answer would be true

Explanation:

Elements 1 through 92 (except for elements 43 and 61) occur naturally on Earth, although some are only present in extremely small quantities.

5 0

2 years ago

3. Balance each of the following redox reactions in the listed aqueous environment. (4 pts each, 8 pts total) Crs)NO3 (a) Cr (a)

GrogVix [38]

Explanation:

(a) The given reaction equation is as follows.

$Cr(s) + NO^{-}_{3}(aq) \rightarrow Cr^{3+}(aq) + NO(g)$ (acidic)

So, here the reduction and oxidation-half reactions will be as follows.

Oxidation-half reaction: $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}$

Reduction-half-reaction: $NO^{-}_{3} + 3e^{-}(aq) \rightarrow NO(g)$

As total charge present on reactant side is -1 and total charge present on product side is +3. And, since it is present in aqueous medium. Hence, we will balance the charge for this reaction equation as follows.

$Cr(s) + NO^{-}_{3}(aq) + 4H^{+}(aq) \rightarrow Cr^{3+}(aq) + NO(g) + 2H_{2}O(l)$ (acidic)

(b) The given reaction equation is as follows.

$HCO^{-}_{3}(aq) + Ag(s) + NH_{3}(aq) \rightarrow H_{2}CO(aq) + Ag(NH_{3})^{+}_{2}(aq)$ (basic)

So, here the reduction and oxidation-half reactions will be as follows.

Reduction-half reaction: $HCO^{-}_{3}(aq) + 4e^{-} \rightarrow H_{2}CO(aq)$

Oxidation-half reaction: $Ag(s) \rightarrow Ag(NH_{3})^{+}_{2}(aq) + 1e^{-}$

Hence, to balance the number of electrons in this equation we multiply it by 4 as follows.

$4Ag(s) \rightarrow 4Ag(NH_{3})^{+}_{2}(aq) + 4e^{-}$

Therefore, balancing the whole reaction equation in the basic medium as follows.

$H_{2}CO(aq) + 4Ag(NH_{3})^{+}_{2}(aq) + 5OH^{-}(aq) \rightarrow HCO^{-}_{3}(aq) + 4Ag(s) + 8NH_{3}(aq) + 3H_{2}O(l)$

6 0

3 years ago