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3 years ago

Which of the following equations has the correct products and is balanced correctly for a reaction between Na3PO4 and KOH?

Chemistry

1 answer:

ASHA 777 [7]3 years ago

5 0

The answer is: A) Na3PO4 + 3KOH → 3NaOH + K3PO4, because K retains the same charge throughout the reaction.

This chemical reaction is double displacement reaction - cations (K⁺ and Na⁺) and anions (PO₄³⁻⁻ and OH⁻) of the two reactants switch places and form two new compounds.

Na₃PO₄ is sodium phosphate.

KOH is potassium hydroxide.

NaOH is sodium hydroxide.

K₃PO₄ is potassium phosphate.

According to the mass conservation law, there are same number of atoms on both side of balanced chemical reaction.

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A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$

and,

$\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903$

Now we have to partial pressure of $C_2HBrClF_3$ and $O_2$ .

According to the Raoult's law,

$p^o=X\times p_T$

where,

$p^o$ = partial pressure of gas

$p_T$ = total pressure of gas

$X$ = mole fraction of gas

$p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T$

$p_{C_2HBrClF_3}=0.0971\times 862torr=84torr$

and,

$p_{O_2}=X_{O_2}\times p_T$

$p_{O_2}=0.903\times 862torr=778torr$

Therefore, the partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

6 0

3 years ago

Give the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by the element. rb

devlian [24]

Actually Rb or Rubidium in zero state has the following electron configuration:

1s22s22p63s23p63d104s24p65s1

However we can see that the ion has a 1 positive charge, which means that it lacks 1 electron, therefore the answer from the choices is:

d. rb+: 1s22s22p63s23p64s23d104p6

3 0

3 years ago

Corinne uses a model rocket in her physical science class. The rocket begins its trip traveling vertically at 22 m s . It reache

dezoksy [38]

The answer is D have a good day

8 0

3 years ago

Read 2 more answers

Indicate the mechanistic symbol (SN1, SN2) that is most consistent with the following statement. Select the single best answer.

Degger [83]

Answer:

The most consistent mechanism is SN1.

Explanation:

The solvolysis of tert-butyl iodide in methanol will arise by the SN1 mechanism.

The reason for that is due the geometric structure of tert-butyl. The tertiary structure allows the formation of a stable carbocation, after the loss of the iodide ion. The carbocation will then be attacked by methanol molecules forming an eter.

4 0

2 years ago

How many grams are in 1.60 moles Hg(NO3)2

Ilia_Sergeevich [38]

Molar mass=MM

MM(Hg)=200.59g/mol
MM(N)=14.01g/mol
MM(O)=16.00g/mol

MM(Hg(NO3)2)=MM(Hg)+2*MM(N)+6*MM(O)=200.59g+2*14.01g+6*16g=324.61g/mol

1.60mol Hg(NO3)2*(324.61g/mol)=519.376 g=519 g Hg(NO3)2

6 0

3 years ago