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Arte-miy333 [17]

3 years ago

7

What is the chemical name of the compound represented by the formula Ni2O3? Use the list of polyatomic ions and the periodic tab

le to help you answer.

2 answers:

bearhunter [10]3 years ago

5 0

The chemical name is nickel (III) oxide .

LuckyWell [14K]3 years ago

4 0

Answer : The chemical name of the compound $Ni_2O_3$ is, nickel(III)oxide.

Explanation :

The rules for naming the chemical name of the compound :

First element in the formula is named first and keep its element name.
Gets a prefix if there is a subscript on it.
Written the oxidation state by the us of roman numeral.
Second element is named second.
Use the root of the element name plus suffix (-ide).
Always use a prefix on the second element.

In the given compound, the nickel is the cation and first element in the formula and the oxidation state of nickel is, (+3). The second element is oxygen which is named as oxide.

Thus, the chemical name of the compound $Ni_2O_3$ is, nickel(III)oxide.

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Zn + O2= ZnO<br><br> How many moles of zinc are needed to make 500. g of zinc oxide?

s2008m [1.1K]

Answer:

5.15 moles

Explanation:

2zn + o2 = 2zno

5.15 2.57 5.15 moles

nzno=500/(16x2+65)= 5.15 moles

-> nzn = 5.15 x 2 ÷ 2 = 5.15 moles

6 0

2 years ago

A laboratory instructor gives a sample of amino acid powder to each of four students, i, ii, iii, and iv, and they weigh the sam

Elena L [17]

Let us differentiate accuracy from precision. Accuracy is the nearness of the measured value to the true or exact value. On the other hand, precision is the nearness of the measured values between each other. So, for precision, select the student in which the measured values are very near to each other. That would be Student III. Now, for accuracy, let's find the average for each student.

Student I: (<span>8.72g+8.74g+8.70g)/3 = 8.72 g
Student II: (</span><span>8.56g+8.77g+8.83g)/3 = 8.72 g
Student III: (</span><span>8.50g+8.48g+8.51g)/3 = 8.50 g
Student IV: (</span><span>8.41g+8.72g+8.55g)/3 = 8.56 g

From the given results, the accurate one would be Students I and II. So, we make a compromise. Even though Student III is precise, it is not accurate. If you compare between Students I and II, the more precise data would be Student I. Therefore, the answer is Student I.</span>

6 0

3 years ago

Read 2 more answers

1Which is a dopant for a p-type semiconductor?

aalyn [17]

For a p type of semiconductor we need a dopant which is from 13th group in periodic table

Al , B, Ga, In Tl

So the correct element will be In : Indium

The other elements belongs to 15th group and hence will give n type semiconductor

5 0

3 years ago

Read 2 more answers

Please answer asap need it by Wednesday morning

yarga [219]

The answer to this problem is quite simple, it’s 9

4 0

3 years ago

Complete and balance the chemical equations for the precipitation reactions, if any, between the following pairs of reactants, a

Crank

Explanation:

a. $Pb(NO_3)_2(aq) + Na_2SO_4(aq)$ → ?

$Pb(NO_3)_2(aq) + Na_2SO_4(aq)\rightarrow PbSO_4(s)+2NaNO_3(aq)$

$Pb(NO_3)_2(aq)\rightarrow Pb^{2+}(aq)+2NO_3^{-}(aq)$

$Na_2SO_4(aq)\rightarrow 2Na^++SO_4^{2-}(aq)$

$Pb^{2+}(aq)+2NO_3^{-}(aq)+2Na^++SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2Na^++2NO_3^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)$

b. $NiCl_2(aq) + NH_4NO_3(aq)$ →

$NiCl_2(aq) + NH4NO_3(aq) \rightarrow Ni(NO_3)_2+NH_4Cl(aq)$

No precipitation is occuring.

c. $Fe_Cl2(aq) + Na_2S(aq)$ →

$FeCl_2(aq) + Na_2S(aq)\rightarrow FeS(s)+2NaCl(aq)$

$FeCl_2(aq)\rightarrow Fe^{2+}(aq)+2Cl^{-}(aq)$

$Na_2S(aq)\rightarrow 2Na^++S{2-}(aq)$

$Fe^{2+}(aq)+2Cl^{-}(aq)+2Na^++S^{2-}(aq)\rightarrow FeS(s)+2Na^++2Cl^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Fe^{2+}(aq)+S^{2-}(aq)\rightarrow FeS(s)$

d. $MgSO_4(aq) + BaCl_2(aq)$ →

$MgSO4(aq) + BaCl2(aq)\rightarrow BaSO_4(s)+MgCl_2$

$MgSO_4(aq)\rightarrow Mg^{2+}(aq)+SO_4^{2-}(aq)$

$BaCl_2(aq)\rightarrow Ba^{2+}+2Cl^{-}(aq)$

$Mg^{2+}(aq)+SO_4^{2-}(aq)+Ba^{2+}+2Cl^{-}(aq)(aq)\rightarrow BaSO_4(s)+Mg^{2+}(aq)+2Cl^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

5 0

2 years ago