Ask question

Ask question

All categories

katovenus [111]

2 years ago

14

A 0.89% (w/v) sodium chloride solution is referred to as physiological saline solution because it has the same concentration of

salts as human blood. what is the molarity of a physiological saline solution? (note: w/v means mass to volume)

1 answer:

maks197457 [2]2 years ago

4 0

1) 0.89% m/v = 0.89 grams of NaCl / 100 ml of solution

=> 8.9 grams of NaCl in 1000 ml of solution = 8.9 grams of NaCl in 1 liter of solution

2) Molarity = M = number of moles of solute / liters of solution

=> calculate the number of moles of 8.9 grams of NaCl

3) molar mass of NaCl = 23.0 g /mol + 35.5 g/mol = 58.5 g / mol

4) number of moles of NaCl = mass / molar mass = 8.9 g / 58.5 g / mol = 0.152 mol

5) M = 0.152 mol NaCl / 1 liter solution = 0.152 M

Answer: 0.152 M

You might be interested in

Use average bond energies to calculate ΔHrxn for the following hydrogenation reaction: H2C=CH2(g)+H2(g)→H3C−CH3(g)

marissa [1.9K]

Answer:

The $\Delta H_{rxn}$ of the given reaction is -129.6 kJ

Explanation:

The given chemical reaction is as follows.

$H_{2}C=CH_{2}(g)+H_{2}(g)\rightarrow H_{3}C-CH_{3}(g)$

Enthalpy of each reactant and products are as follows.

$\Delta H_{C=C}\,=615.0\,kJ\,mol^{-1}$

$\Delta H _{H-H}\,=435.1\,kJ\,mol^{-1}$

$\Delta H _{C-C}\,=347.3\,kJ\,mol^{-1}$

$\Delta H _{C-H}\,=416.2\,kJ\,mol^{-1}$

In the given chemical reaction involved two C-H bonds in the reactant side and one C-C bond in the product side therefore, the enthalpy of formation will be the negative.

$\Delta H_{rxn}=-\Delta H_{C-C}-2\Delta H_{C-H}+\Delta H_{C=C}+\Delta H_{H-H}$

$=-347.4-2\times416.2+615.0+435.1$

$=-129.6 \,kJ$

Therefore, The $\Delta H_{rxn}$ of the given reaction is -129.6 kJ

4 0

2 years ago

What characterizes an Arrhenius acid

dedylja [7]

The characteristics of an Arrhenius acid was the acid donates H+ when it is in the aqueous solution, whether the solution is acid or base. If the solution is a acid, then be H+ builds up and increase its thing. The pH level for acid solution is the scale from 1-6. Even though the ph scale is lower than the base, the color also varies. It also applies to the base, too.

8 0

2 years ago

At 4.00 L, an expandable vessel contains 0.864 mol of oxygen gas. How many liters of oxygen gas must be added at constant temper

fgiga [73]

Answer:

We have to add 2.30 L of oxygen gas

Explanation:

Step 1: Data given

Initial volume = 4.00 L

Number of moles oxygen gas= 0.864 moles

Temperature = constant

Number of moles of oxygen gas increased to 1.36 moles

Step 2: Calculate new volume

V1/n1 = V2/n2

⇒V1 = the initial volume of the vessel = 4.00 L

⇒n1 = the initial number of moles oxygen gas = 0.864 moles

⇒V2 = the nex volume of the vessel

⇒n2 = the increased number of moles oxygen gas = 1.36 moles

4.00L / 0.864 moles = V2 / 1.36 moles

V2 = 6.30 L

The new volume is 6.30 L

Step 3: Calculate the amount of oxygen gas we have to add

6.30 - 4.00 = 2.30 L

We have to add 2.30 L of oxygen gas

4 0

2 years ago

The density of whole blood is 1.05 g/ml. a typical adult has between 4.7 and 5.5 l of whole blood. what is the mass in pounds of

Kisachek [45]

<span>1.05 g/ml * 1000 ml = 1050g/l because of 1g/ml = 1 kg/l so, a/q mass of 4.7 l of whole blood in pound = 4.7 * 1050 = 4935 g so in pound 4935g = 10.87981p</span>

5 0

2 years ago

Determine net ionic equations, if any, occuring when aqueous solutions of the following reactants are mixed. Select "True" or "F

nirvana33 [79]

<u>Answer:</u>

<u>For 1:</u> The correct answer is False.

<u>For 2:</u> The correct answer is True.

<u>For 3:</u> The correct answer is True.

<u>For 4:</u> The correct answer is False.

<u>For 5:</u> The correct answer is True.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions. If no net ionic equation is formed, it is said that no reaction has occurred.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form. Solids, liquids and gases do not exist as ions.

<u>For 1:</u> Lead(II) nitrate and sodium chloride

The chemical equation for the reaction of lead (II) nitrate and sodium chloride is given as:

$Pb(NO_3)_2(aq.)+2NaCl(aq.)\rightarrow PbCl_2(s)+2NaNO_3(aq.)$

Ionic form of the above equation follows:

$Pb^{2+}(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+2Na^+(aq.)+2NO_3^-(aq.)$

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)$

Hence, the correct answer is False.

<u>For 2:</u> Sodium bromide and hydrochloric acid

The chemical equation for the reaction of sodium bromide and hydrochloric acid is given as:

$NaBr(aq.)+HCl(aq.)\rightarrow NaCl(aq.)+HBr(aq.)$

Ionic form of the above equation follows:

$Na^{+}(aq.)+Br^-(aq.)+H^+(aq.)+Cl^-(aq.)\rightarrow Na^+(aq.)+Cl^-(aq.)+H^+(aq.)+Br^-(aq.)$

There are no spectator ions in the equation. So, the above reaction is the net ionic equation.

Hence, the correct answer is True.

<u>For 3:</u> Nickel (II) chloride and lead(II) nitrate

The chemical equation for the reaction of lead (II) nitrate and nickel (II) chloride is given as:

$Pb(NO_3)_2(aq.)+NiCl_2(aq.)\rightarrow PbCl_2(s)+Ni(NO_3)_2(aq.)$

Ionic form of the above equation follows:

$Pb^{2+}(aq.)+2NO_3^-(aq.)+Ni^{2+}(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+Ni^{2+}(aq.)+2NO_3^-(aq.)$

As, nickel and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)$

Hence, the correct answer is True.

<u>For 4:</u> Magnesium chloride and sodium hydroxide

The chemical equation for the reaction of magnesium chloride and sodium hydroxide is given as:

$MgCl_2(aq.)+2NaOH(aq.)\rightarrow Mg(OH)_2(s)+2NaCl(aq.)$

Ionic form of the above equation follows:

$Mg^{2+}(aq.)+2Cl^-(aq.)+2Na^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2Na^+(aq.)+2Cl^-(aq.)$

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Mg^{2+}(aq.)+OH^-(aq.)\rightarrow Mg(OH)_2(s)$

Hence, the correct answer is False.

<u>For 5:</u> Ammonium sulfate and barium nitrate

The chemical equation for the reaction of ammonium sulfate and barium nitrate is given as:

$(NH_4)_2SO_4(aq.)+Ba(NO_3)_2(aq.)\rightarrow BaSO_4(s)+2NH_4NO_3(aq.)$

Ionic form of the above equation follows:

$2NH_4^{+}(aq.)+SO_4^{2-}(aq.)+Ba^{2+}(aq.)+2NO_3^-(aq.)\rightarrow BaSO_4(s)+2NH_4^+(aq.)+2NO_3^-(aq.)$

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)$

Hence, the correct answer is True.

3 0

2 years ago