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AURORKA [14]
2 years ago
14

Titanium has five common isotopes: 46Ti (8.0%), 47Ti (7.8%), 48Ti (73.4%), 49Ti (5.5%), 50 Ti (5.3%). What is the average atomic

mass of titanium?
Chemistry
1 answer:
zhannawk [14.2K]2 years ago
6 0
(46x8.0)+(47x7.8)+(48x73.4)+(49x5.5)+(50x5.3) = 4792.3

4792.3/100 = 47.923 this is the average atomic mass of Titanium
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Determine the compound type for the following formulas: C12H22011 Mg(OH)2 H20 Cu3Zn2 Au <br>​
Scorpion4ik [409]

Answer:

C12H22O11  

✔ covalent

Mg(OH)2    

✔ Ionic

H2O    

✔ covalent

Cu3Zn2    

✔ metallic

Au      

✔ metallic

Explanation:

7 0
2 years ago
Read 2 more answers
II. Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.86 g of magnesium ribbo
Leto [7]

Answer:

Excess=3.53g

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2Mg(s)+O_2(g) \rightarrow 2MgO(s)

Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:

n_{MnO}^{by \ Mg}=3.86gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg}  =0.159molMgO\\\\n_{MnO}^{by \ O_2}=\frac{1atm*0.155L}{0.082\frac{atm*L}{molO_2*K}*275K} *\frac{2mol MgO}{1molO_2} =0.0137molMgO

It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:

m_{Mg}^{consumed}=0.0137molMgO*\frac{2molMg}{2molMgO} *\frac{24.3gMg}{1molMg} =0.334gMg

Thus, the mass in excess is:

Excess=3.86g-0.334g\\\\Excess=3.53g

Regards!

5 0
1 year ago
Determine whether each description applies to electrophilic aromatic substitution or nucleophilic aromatic substitution.
Alborosie

Answer:

a. electrophilic aromatic substitution

b. nucleophilic aromatic substitution

c. nucleophilic aromatic substitution

d. electrophilic aromatic substitution

e. nucleophilic aromatic substitution

f. electrophilic aromatic substitution

Explanation:

Electrophilic aromatic substitution is a type of chemical reaction where a hydrogen atom or a functional group that is attached to the aromatic ring is replaced by an electrophile. Electrophilic aromatic substitutions can be classified into five classes: 1-Halogenation: is the replacement of one or more hydrogen (H) atoms in an organic compound by a halogen such as, for example, bromine (bromination), chlorine (chlorination), etc; 2- Nitration: the replacement of H with a nitrate group (NO2); 3-Sulfonation: the replacement of H with a bisulfite (SO3H); 4-Friedel-CraftsAlkylation: the replacement of H with an alkyl group (R), and 5-Friedel-Crafts Acylation: the replacement of H with an acyl group (RCO). For example, the Benzene undergoes electrophilic substitution to produce a wide range of chemical compounds (chlorobenzene, nitrobenzene, benzene sulfonic acid, etc).

A nucleophilic aromatic substitution is a type of chemical reaction where an electron-rich nucleophile displaces a leaving group (for example, a halide on the aromatic ring). There are six types of nucleophilic substitution mechanisms: 1-the SNAr (addition-elimination) mechanism, whose name is due to the Hughes-Ingold symbol ''SN' and a unimolecular mechanism; 2-the SN1 reaction that produces diazonium salts 3-the benzyne mechanism that produce highly reactive species (including benzyne) derived from the aromatic ring by the replacement of two substituents; 4-the free radical SRN1 mechanism where a substituent on the aromatic ring is displaced by a nucleophile with the formation of intermediary free radical species; 5-the ANRORC (Addition of the Nucleophile, Ring Opening, and Ring Closure) mechanism, involved in reactions of metal amide nucleophiles and substituted pyrimidines; and 6-the Vicarious nucleophilic substitution, where a nucleophile displaces an H atom on the aromatic ring but without leaving groups (such as, for example, halogen substituents).

3 0
2 years ago
For the reaction n2(g) + 2h2(g) â n2h4(l), if the percent yield for this reaction is 77.5%, what is the actual mass of hydrazine
Rudiy27

First calculate the moles of N2 and H2 reacted.

moles N2 = 27.7 g / (28 g/mol) = 0.9893 mol

moles H2 = 4.45 g / (2 g/mol) = 2.225 mol

 

We can see that N2 is the limiting reactant, therefore we base our calculation from that.

Calculating for mass of N2H4 formed:

mass N2H4 = 0.9893 mol N2 * (1 mole N2H4 / 1 mole N2) * 32 g / mol * 0.775

<span>mass N2H4 = 24.53 grams</span>

7 0
2 years ago
In an experiment, 0.42 mol of co and 0.42 mol of h2 were placed in a 1.00-l reaction vessel. at equilibrium, there were 0.29 mol
Likurg_2 [28]
To determine the Keq, we need the chemical reaction in the system. In this case it would be:

CO + 2H2 = CH3OH

The Keq is the ration of the amount of the product and the reactant. We use the ICE table for this. We do as follows:

          CO           H2             CH3OH
I         .42           .42                    0
C     -0.13      -2(0.13)            0.13
-----------------------------------------------
E =    .29           0.16               0.13

Therefore, 

Keq = [CH3OH] / [CO2] [H2]^2 = 0.13 / 0.29 (0.16^2)
Keq = 17.51
4 0
2 years ago
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