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3 years ago

Which of the following polyatomic ions will form an ionic compound with a single sodium ion? CO32− PO43− SO42− NO31−

Chemistry

1 answer:

Artist 52 [7]3 years ago

4 0

A single sodium ion will combine due to its charge on it

Na^+

The single sodium ion has total one positive charge on it. so it requires only one electron or can bond to one negative charge

in the given polyatomic anions

a) CO3^-2 : it has two negative charge. So it will react with two sodium ions

it will forms Na2CO3

b) PO4^-3 :it has three negative charge. So it will react with three sodium ions

it will forms Na3PO4

c) SO4^-2 : it has two negative charge. So it will react with two sodium ions

it will forms Na2SO4

d) NO3^-1: it has one negative charge. So it will react with one sodium ion

it will forms NaNO3

Hence the correct answer is

Nitrate ion will react with single sodium ion as

Na+ + NO3- ---> NaNO3

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A chemical reaction is shown below:

BabaBlast [244]

Answer:

Mass = 8.46 g

Explanation:

Given data:

Mass of water produced = ?

Mass of glucose = 20 g

Mass of oxygen = 15 g

Solution:

Chemical equation:

C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂

Number of moles of glucose:

Number of moles = mass/molar mass

Number of moles = 20 g/ 180.16 g/mol

Number of moles = 0.11 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 15 g/ 32 g/mol

Number of moles = 0.47 mol

now we will compare the moles of water with oxygen and glucose.

C₆H₁₂O₆ : H₂O

1 : 6

0.11 : 6/1×0.11 = 0.66

O₂ : H₂O

6 : 6

0.47 : 0.47

Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 0.47 mol ×18 g/mol

Mass = 8.46 g

8 0

3 years ago

Please help! Don’t know if I’m right!!

kompoz [17]

Answer:e

Explanation:

Bananas

4 0

3 years ago

A hydrate is typically a crystalline compound in which water molecules are chemically bound to another compound or an element.

dedylja [7]

Answer:

Subtract the mass of the CuSO4⋅ 5H2O from the mass of CuSO4 is the right one

7 0

3 years ago

An independent variable is:

weqwewe [10]

Answer: The variable that you change in an experiment

Explanation: It does not rely on any other factors that deal with the experiment

6 0

3 years ago

If 45.00 g of precipitate is formed from the reaction of 0.100 mol/L

Tom [10]

Answer:

Approximately $2.53\; \rm L$ (rounded to three significant figures) assuming that ${\rm HCl}\, (aq)$ is in excess.

Explanation:

When ${\rm HCl} \, (aq)$ and ${\rm AgNO_3}\, (aq)$ precipitate, ${\rm AgCl} \, (s)$ (the said precipitate) and $\rm HNO_3\, (aq)$ are produced:

${\rm HCl}\, (aq) + {\rm AgNO_3}\, (aq) \to {\rm AgCl}\, (s) + {\rm HNO_3}\, (aq)$ (verify that this equation is indeed balanced.)

Look up the relative atomic mass of $\rm Ag$ and $\rm Cl$ on a modern periodic table:

$\rm Ag$ : $107.868$ .
$\rm Cl$ : $35.45$ .

Calculate the formula mass of the precipitate, $\rm AgCl$ :

$\begin{aligned}& M({\rm AgCl})\\ &= (107.868 + 35.45)\; \rm g \cdot mol^{-1} \\\ &\approx 143.318 \; \rm g\cdot mol^{-1}\end{aligned}$ .

Calculate the number of moles of $\rm AgCl$ formula units in $45.00\; \rm g$ of this compound:

$\begin{aligned}n({\rm AgCl}) &= \frac{m({\rm AgCl})}{M({\rm AgCl})} \\ &\approx \frac{45.00\; \rm g}{143.318\; \rm g \cdot mol^{-1}}\approx 0.313987\; \rm mol \end{aligned}$ .

Notice that in the balanced equation for this reaction, the coefficients of ${\rm AgNO_3} \, (aq)$ and ${\rm AgCl}\, (s)$ are both one.

In other words, if ${\rm HCl}\, (aq)$ (the other reactant) is in excess, it would take exactly $1\; \rm mol$ of ${\rm AgNO_3} \, (aq)\!$ formula units to produce $1\; \rm mol \!$ of ${\rm AgCl}\, (s)\!$ formula units.

Hence, it would take $0.313987\; \rm mol$ of ${\rm AgNO_3} \, (aq)\!$ formula units to produce $0.313987\; \rm mol\!$ of ${\rm AgCl}\, (s)\!$ formula units.

Calculate the volume of the ${\rm AgNO_3} \, (aq)\!$ solution given that the concentration of the solution is $0.124\; \rm mol \cdot L^{-1}$ :

$\begin{aligned}V({\rm AgNO_3}) &= \frac{n({\rm AgNO_3})}{c({\rm AgNO_3})} \\ &\approx \frac{0.313987\; \rm mol}{0.124\; \rm mol \cdot L^{-1}}\approx 2.53\; \rm L\end{aligned}$ .

(The answer was rounded to three significant figures so as to match the number of significant figures in the concentration of ${\rm AgNO_3} \, (aq)\!$ .)

In other words, approximately $2.53\; \rm L$ of that ${\rm AgNO_3} \, (aq)\!$ solution would be required.

4 0

3 years ago