How many neutrons does carbon 16 have.

When potassium hydroxide and barium chloride react, potassium chloride and barium hydroxide are formed. The balanced equation fo

Mashcka [7]

Answer:

The answer to your question is letter C.

Explanation:

Reaction

Potassium hydroxide = KOH

Barium chloride = BaCl₂

Potassium chloride = KCl

Barium hydroxide = Ba(OH)₂

KOH + BaCl₂ ⇒ KCl + Ba(OH)₂

Reactant Elements Products

1 K 1

1 Ba 1

2 Cl 1

1 H 2

1 O 2

The reaction is unbalanced

2KOH + BaCl₂ ⇒ 2KCl + Ba(OH)₂

Reactant Elements Products

2 K 2

1 Ba 1

2 Cl 2

2 H 2

2 O 2

Now, the reaction is balanced

7 0

3 years ago

Read 2 more answers

How many doubtful digit(s) is/are allowed in any measured quantity?

Mnenie [13.5K]

Answer:

zero

Explanation:

I I think one should be so accurate with measurements and experiments

3 0

3 years ago

Help me answer this question for points!

labwork [276]

Answer:

I am pretty sure Danny Duncan told me 69

Explanation:

niice

6 0

3 years ago

Given the following reaction: 2K3PO4 + AL2(CO3)3 = 3K2CO3 + 2ALPO4 If I perform this reaction with 150 g of potassium phosphate

Novosadov [1.4K]

Answer : The theoretical yield of potassium carbonate is, 146.483 g

The percent yield of potassium carbonate is, 85.33 %

Solution : Given,

Mass of $K_3PO_4$ = 150 g

Mass of $Al_2(CO_3)_3$ = 90 g

Molar mass of $K_3PO_4$ = 212.27 g/mole

Molar mass of $Al_2(CO_3)_3$ = 233.99 g/mole

Molar mass of $K_2CO_3$ = 138.205 g/mole

First we have to calculate the moles of $K_3PO_4$ and $Al_2(CO_3)_3$

$\text{ Moles of }K_3PO_4=\frac{\text{ Mass of }K_3PO_4}{\text{ Molar mass of }K_3PO_4}=\frac{150g}{212.27g/mole}=0.7066moles$

$\text{ Moles of }Al_2(CO_3)_3=\frac{\text{ Mass of }Al_2(CO_3)_3}{\text{ Molar mass of }Al_2(CO_3)_3}=\frac{90g}{233.99g/mole}=0.3846moles$

The given balanced reaction is,

$2K_3PO_4+Al_2(CO_3)_3\rightarrow 3K_2CO_3+2AlPO_4$

From the given reaction, we conclude that

2 moles of $K_3PO_4$ react with 1 mole of $Al_2(CO_3)_3$

0.7066 moles of $K_3PO_4$ react with $\frac{1}{2}\times 0.7066=0.3533$ moles of $Al_2(CO_3)_3$

But the moles of $Al_2(CO_3)_3$ is, 0.3846 moles.

So, $Al_2(CO_3)_3$ is an excess reagent and $K_3PO_4$ is a limiting reagent.

Now we have to calculate the moles of $K_2CO_3$ .

As, 2 moles of $K_3PO_4$ react to give 3 moles of $K_2CO_3$

So, 0.7066 moles of $K_3PO_4$ react to give $\frac{3}{2}\times 0.7066=1.0599$ moles of $K_2CO_3$

Now we have to calculate the mass of $K_2CO_3$ .

$\text{ Mass of }K_2CO_3=\text{ Moles of }K_2CO_3\times \text{ Molar mass of }K_2CO_3$

$\text{ Mass of }K_2CO_3=(1.0599moles)\times (138.205g/mole)=146.483g$

The theoretical yield of potassium carbonate = 146.483 g

The experimental yield of potassium carbonate = 125 g

Now we have to calculate the % yield of potassium carbonate.

Formula for percent yield :

$\% yield=\frac{\text{ Theoretical yield}}{\text{ Experimental yield}}\times 100$

$\% \text{ yield of }K_2CO_3=\frac{125g}{146.483g}\times 100=85.33\%$

Therefore, the % yield of potassium carbonate is, 85.33%

6 0

3 years ago

Match the prefix used in naming binary molecule compounds with its corresponding number

VARVARA [1.3K]

Nona= 9, hepta= 7, hexa= 6, tetra= 4

8 0

3 years ago