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2 years ago

Blank Na2CO3 + blank Ag3P --> blank Na3P + blank Ag2CO3

Chemistry

1 answer:

mezya [45]2 years ago

8 0

Answer: __3__Na2CO3 + __2__Ag3P → __2__Na3P + __3__Ag2CO3

Explanation:

In balancing equations you need to make sure the number of atoms before the reaction should be equal to the number of atoms after the reaction.

Before the reaction: We had 6Na, 9CO, 6Ag and 2p

After the reaction: We have 6Na, 9CO, 6Ag and 2p

And now we are sure that atoms before the reaction are equal to atoms after the reaction

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Givin 11 sodium 23, what is the atomic mass

shutvik [7]

The atomic mass of sodium is 23. i hope that helps!

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2 years ago

What problems might happen if an unnecessary hurricane warning was issues

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People would rush to the store to buy supplies and there might not be enough for the "last minute noticed" people

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3 years ago

Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu

pochemuha

Answer : The volume of $CO_2$ will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

$HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2$

First we have to calculate the mass of $HCO_3^-$ in tablet.

$\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g$

Now we have to calculate the moles of $HCO_3^-$ .

Molar mass of $HCO_3^-$ = 1 + 12 + 3(16) = 61 g/mole

$\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles$

Now we have to calculate the moles of $CO_2$ .

From the balanced chemical reaction, we conclude that

As, 1 mole of $HCO_3^-$ react to give 1 mole of $CO_2$

So, 0.0202 mole of $HCO_3^-$ react to give 0.0202 mole of $CO_2$

The moles of $CO_2$ = 0.0202 mole

Now we have to calculate the volume of $CO_2$ by using ideal gas equation.

$PV=nRT$

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = $37^oC=273+37=310K$

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

$(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)$

$V=0.51411L=514.11ml$

Therefore, the volume of $CO_2$ will be, 514.11 ml

6 0

2 years ago

How many moles of phosphoric acid would be needed to produce 15 grams of water?

bekas [8.4K]

Moles of phosphoric acid would be needed : 0.833

<h3>Further explanation</h3>

Given

15 grams of water

Required

moles of phosphoric acid

Solution

Reaction(decomposition) :

H3PO4 -> H2O + HPO3

mol water (H2O :

= mass : MW

= 15 g : 18 g/mol

= 0.833

From the equation, mol ratio H3PO4 = mol H2O = 1 : 1, so mol H3PO4 = 0.833

7 0

2 years ago

I need you to solve for me plzzzzzz

Sergeu [11.5K]

Answer:

.0556 L

Explanation:

First, convert the 1.35 M to 1.35 mol/L in order for the units to correctly cancel out.

Then, multiply (0.0725 moles Na2CO3/1) times (L/ 1.35 mol).

Finally, the answer will be .0556 L.

<h3 />

3 0

2 years ago