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3 years ago

What is the formula of titanium(IV) bromide?

Chemistry

1 answer:

Anna [14]3 years ago

3 0

Answer:

$\boxed{\text{SnBr}_{4}}$

Explanation:

The name tells you that this is a binary compound (contains two elements).

It contains a metal and a nonmetal, so it is a binary ionic compound. The general rule is:

Name of compound = name of metal name of ion (two words)

Name of metal = tin(IV), so the tin ion has a charge of 4+

Name of ion = bromide. Br is in Group 17, so bromide ion has charge of 1-.

$\rm Sn$^{4+}$ + 4Br$^{-} \longrightarrow \,$ SnBr$_{4}$\\\\\text{The formula is} \boxed{\textbf{SnBr}_{4}}$

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The trait for flower color in a plant has red and white alleles. The red color is the dominant trait. What is the phenotypic rat

Fudgin [204]

Answer :

2 red : 2 white

Explanation:

;)long story short

hetro with hetro gives 3:1
hetro with recessive gives 1:1

Hetro :- ( Rr) one capital letter and one small
Recessive :- (rr) two small leters

6 0

3 years ago

A student prepares 150.0 mL of 1.40 M HCl using 35.0 mL of a stock solution. What is the concentration of the stock solution? Us

drek231 [11]

Answer:6.0M

Explanation:see attached photo

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4 years ago

What is the gram formula of na2co3

makkiz [27]

Answer:

106 gfm

Explanation:

element. number masses

Na. 2 23×2

C. 1. 12

O. 3. 16×3

then add

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5 0

3 years ago

How many molecules of sulfur trioxide are in 78.0 grams?

zalisa [80]

Answer:

b) 5.87 E23 molecules

Explanation:

∴ mm SO3 = 80.066 g/mol

⇒ molecules SO3 = (78.0 g)(mol/80.066 g)(6.022 E23 molec/mol)

⇒ molec SO3 = 5.866 E23 molecules SO3

5 0

3 years ago

An iron ore sample weighing 0.5562 g is dissolved HCl (aq), and the iron is obtained as Fe2 in solution. This solution is then t

erik [133]

Answer:

$\% Fe^{+2}=70%$

Explanation:

Hello,

In this case, we could considering this as a redox titration:

$Fe^{+2}+(Cr_2O_7)^{-2}\rightarrow Fe^{+3}+Cr^{+3}$

Thus, the balance turns out (by adding both hydrogen ions and water):

$Fe^{+2}\rightarrow Fe^{+3}+1e^-\\(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\6Fe^{+2}+(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow Cr^{+3}+7H_2O+6Fe^{+3}+6e^-$

Thus, by stoichiometry, the grams of Fe+2 ions result:

$m_{Fe^{+2}}=0.04021\frac{molK_2Cr_2O_7}{L}*0.02872L*\frac{6molFe^{+2}}{1molK_2Cr_2O_7}*\frac{56gFe^{+2}}{1molFe^{+2}}=0.388gFe^{+2}$

Finally, the mass percent is:

$\% Fe^{+2}=\frac{0.388g}{0.5562g}*100\%\\ \% Fe^{+2}=70%$

Best regards.

8 0

3 years ago