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2 years ago

Place the following in order of increasing F-A-F bond angle, where A represents the central atom in each molecule. PF3 OF2 PF4

Chemistry

1 answer:

iren2701 [21]2 years ago

8 0

Answer:

Correct option : B

Explanation:

First compare $PF_3$ with $OF_2$

there are two lone pair in $OF_2$ but only one lone pair presen in the $PF_3$ so more lone pair means more repulsion and hence less bond bond angle beacuse of lone pair repulsion bonded atom come closer and hence bond angle decreased .

bond angle inversely proporional to repulsion so more repulsion means less bond angle

$bond \, angle \, : OF_2 \textless PF_3$

$PF_4$ does no exist it exist only in the form of $PF_4^+ \, or\, PF_4^-$ and here assuming $PF_4^+$ because in case of $PF_4^-$ difficult to compare bond angle because here two different type of bond angle is present.

$Compare \, PF_3 \, and \, PF_4^+$

there is no lone pair in the $PF_4^+$ so here will no be any repulsion and hence bond will be more in $PF_4^+$ as compared to $PF_3$

$bond \, angle \, : PF_3 \textless PF_4^+$

$bond \, angle \, : OF_2 \textless PF_3 \textless PF_4^+$

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Which of the following sets of quantum numbers is NOT allowed? a. n = 5, l= 4, ml= –2, ms = +1/2 b. n = 2, l = 1, ml= 0, ms = +1

klemol [59]

Answer:

Explanation:

the n value must always be greater than the l value

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3 years ago

Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)

cupoosta [38]

Answer:

$m_{CaCO_3}=0.179gCaCO_3$

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

$PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2$

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

$m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3$

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3 0

3 years ago

What is the mass of 2.23 × 10^23 atoms of aluminum?

rjkz [21]

Answer:

Explanation:

In image

Take atomic mass or molar mass

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7 0

2 years ago

2074 Set B Q.No. 1 What mass of nitrogen will be requires

timurjin [86]

140 g of nitrogen (N₂)

Explanation:

We have the following chemical equation:

N₂ + 3 H₂ -- > 2 NH₃

Now, to find the number of moles of ammonia we use the Avogadro's number:

if 1 mole of ammonia contains 6.022 × 10²³ molecules

then X moles of ammonia contains 6.022 × 10²⁴ molecules

X = (1 × 6.022 × 10²⁴) / 6.022 × 10²³

X = 10 moles of ammonia

Taking in account the chemical reaction we devise the following reasoning:

If 1 mole of nitrogen produces 2 moles of ammonia

then Y moles of nitrogen produces 10 moles of ammonia

Y = (1 × 10) / 2

Y = 5 moles of nitrogen

number of moles = mass / molecular weight

mass = number of moles × molecular weight

mass of nitrogen (N₂) = 5 × 28 = 140 g

Learn more about:

Avogadro's number

brainly.com/question/13772315

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3 years ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

3 years ago