Descuss how to find the oldest paper in a stack of papers

mario62 [17]

Answer:

Complete ionic: $\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Net ionic: $\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

Explanation:

Start by identifying species that exist as ions. In general, such species include:

Soluble salts.
Strong acids and strong bases.

All four species in this particular question are salts. However, only three of them are generally soluble in water: $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ . These three salts will exist as ions:

Each $\rm AgNO_3\, (aq)$ formula unit will exist as one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion.
Each $\rm CaCl_2$ formula unit will exist as one $\rm Ca^{2+}$ ion and two $\rm Cl^{-}$ ions (note the subscript in the formula $\rm CaCl_2\!$ .)
Each $\rm Ca(NO_3)_2$ formula unit will exist as one $\rm Ca^{2+}$ and two $\rm {NO_3}^{-}$ ions.

On the other hand, $\rm AgCl$ is generally insoluble in water. This salt will not form ions.

Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ (three soluble salts) as the corresponding ions.

Pay attention to the coefficient of each species. For example, indeed each $\rm AgNO_3\, (aq)$ formula unit will exist as only one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion. However, because the coefficient of $\rm AgNO_3\, (aq)\!$ in the original equation is two, $\!\rm AgNO_3\, (aq)$ alone should correspond to two $\rm Ag^{+}\!$ ions and two $\rm {NO_3}^{-}\!$ ions.

Do not rewrite the salt $\rm AgCl$ because it is insoluble.

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of $\rm Ca^{2+}$ and two units of $\rm {NO_3}^{-}$ . Doing so will give:

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, Cl^{-}\, (aq) \to 2\, AgCl\, (s)\end{aligned}$ .

Simplify the coefficients:

$\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

7 0

2 years ago

A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h

statuscvo [17]

Explanation:

First, calculate the moles of $CO_{2}$ using ideal gas equation as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}$ (as 1 bar = 1 atm (approx))

= 0.183 mol

As, Density = $\frac{mass}{volume}$

Hence, mass of water will be as follows.

Density = $\frac{mass}{volume}$

0.998 g/ml = $\frac{mass}{3.26 ml}$

mass = 3.25 g

Similarly, calculate the moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{3.25 g}{18.02 g/mol}$

= 0.180 mol

Moles of hydrogen = $0.180 \times 2$ = 0.36 mol

Now, mass of carbon will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.183 mol = $\frac{mass}{12 g/mol}$

= 2.19 g

Therefore, mass of oxygen will be as follows.

Mass of O = mass of sample - (mass of C + mass of H)

= 3.50 g - (2.19 g + 0.36 g)

= 0.95 g

Therefore, moles of oxygen will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.95 g}{16 g/mol}$

= 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

C H O

No. of moles: 0.183 0.36 0.059

On dividing: 3.1 6.1 1

Therefore, empirical formula of the given compound is $C_{3}H_{6}O$ .

Thus, we can conclude that empirical formula of the given compound is $C_{3}H_{6}O$ .

6 0

3 years ago

You are on the Titanic and want to find something that will float because you didn't

damaskus [11]

Find volume of pillow

L=78cm

B=55cm

H=25cm

$\\ \bull\tt\longrightarrow V=LBH$

$\\ \bull\tt\longrightarrow V=25(78)(55)$

$\\ \bull\tt\longrightarrow V=107250cm^3$

$\\ \bull\tt\longrightarrow V=10.72m^3$

Now

Mass=5.5kg

$\\ \bull\tt\longrightarrow Density=\dfrac{Mass}{Volume}$

$\\ \bull\tt\longrightarrow Density=\dfrac{5.5}{10.72}$

$\\ \bull\tt\longrightarrow Density=0.5kg/m^3$

Density of water=1000kg/m^3

As it is less than density of water it will float on water

8 0

2 years ago

What does M stand for

liq [111]

Answer:Molarity

Explanation:M stand for molarity

6 0

3 years ago

what is the mass of water vapor produced when 3.2 liters reacts with 8.7 liters of oxygen gas at STP?

uranmaximum [27]

Answer:

2.57g

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

2H2 + O2 —> 2H2O

Next let us determine the limiting reactant. This is achieved as follows:

From the equation,

2L H2 required 1L of O2.

Therefore, 3.2L of H will require = 3.2/2 = 1.6L of O2

From the calculation above, O2 is excess because the volume of O2 given from the question is far greater than the volume of O2 obtained from our calculation. Therefore, H2 is the limiting reactant.

Now let us covert 3.2L of H2 to mole. This is illustrated below:

1mole of a gas occupy 22.4L at stp

Therefore, Xmol of H2 will occupy 3.2L i.e

Xmol of H2 = 3.2/22.4 = 0.143mol

From the equation,

2moles of H2 produced 2moles of H2O.

Therefore, 0.143mol of H2 will also produce 0.143moles of H2O.

Now, we can obtain the mass of the water vapour produced by convert 0.143mol of H2O to gram. This is illustrated below:

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Number of mole of H2O = 0.143mol

Mass of H2O =?

Mass = mole x Molar Mass

Mass of H2O = 0.143 x 18 = 2.57g

The mass of water vapour produce is 2.57g

8 0

2 years ago