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2 years ago

Which of the following lists describes characteristics of a base?

Chemistry

2 answers:

soldi70 [24.7K]2 years ago

5 0

Answer: Bitter taste, high pH, and caustic

Explanation:

Acids are those substances which either donates hydrogen ions $(H^+)$ when dissolved in water. They have pH ranging from 1 to 6.9.They are They are sour in taste. They dissolve metals to give hydrogen gas.Strong acids are caustic as they burn skin.

$HX\rightarrow H^+X^-$

Bases are those substances which either donates hydroxide ions $(OH^-)$ when dissolved in water or donates a pair of electrons. They have pH ranging from 7.1 to 14. They are bitter in taste. They are slippery in nature. Strong bases are caustic as they burn skin.

$BOH\rightarrow B^++OH^-$

suter [353]2 years ago

3 0

Answer:

bitter taste low ph and slipery

Explanation:

if its sour or dissolves metal it is an acid

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kondor19780726 [428]

Explanation:

It is known that 1 gram contains 1000 milligrams. And, mathematically we can represent it as follows.

$\frac{1 g}{1000 mg}$ or $\frac{1000 mg}{1 g}$

So, when we have to convert grams into milligrams then we simply multiply the digit with 1000. And, if we have to convert a digit from milligrams to grams then we simply divide it by 1000.

4 0

2 years ago

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit

Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

$x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L$

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0

2 years ago

What mass (g) of barium iodide is contained in 188 ml of a barium iodide solution that has an iodide ion concentration of 0.532m

Katarina [22]

Answer:

What mass (g) of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?

A) 19.6

B) 39.1

C) 19,600

D) 39,100

E) 276

The correct answer to the question is

B) 39.1 grams

Explanation:

To solve the question

The molarity ratio is given by

188 ml of 0.532 M solution of iodide.

Therefore we have number of moles = 0.188 × 0.532 M = 0.100016 Moles

To find the mass, we note that the Number of moles = $\frac{Mass}{Molar Mass}$ from which we have

Mass = Number of moles × molar mass

Where the molar mass of Barium Iodide = 391.136 g/mol

= 0.100016 moles ×391.136 g/mol = 39.12 g

8 0

2 years ago

A hypothetical AX type of ceramic material is known to have a density of 3.55 g/cm3 and a unit cell of cubic symmetry with a cel

postnew [5]

Explanation:

For AX type ceramic material, the number of formula per unit cells is as follows.

$\rho = \frac{n'(A_{c} + A_{A})}{V_{C}N_{A}}$

or, $n' = \frac{\rho a^{3}N_{A}}{(A_{c} + A_{A})}$

where, n' = no. of formula units per cell

$A_{c}$ = molecular weight of cation = 90.5 g/mol

$A_{a}$ = molecular weight of anion = 37.3 g/mol

$V_{c}$ = volume of cubic cell = 3.55 $g/cm^{3}$

a = edge length of unit cell = $3.9 \times 10^{-8} cm$

$N_{A}$ = Avogadro's number = $6.023 \times 10^{23}$

$\rho$ = density = 3.55 $g/cm^{3}$

Now, putting the given values into the above formula as follows.

$n' = \frac{\rho a^{3}N_{A}}{(A_{c} + A_{A})}$

= $\frac{3.55 g/cm^{3} \times (3.9 \times 10^{-8})^{3} \times 6.023 \times 10^{23}}{(90.5 + 37.3)}$

= 0.9

= 1 (approx)

Therefore, we can conclude that out of the given options crystal structure of cesium chloride is possible for this material.

3 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]

Answer:

For 1: The correct answer is False.

For 2: The correct answer is True.

For 3: The correct answer is True.

For 4: The correct answer is False.

For 5: The correct answer is True.

Explanation:

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.

For 1: 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.

For 2: 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.

For 3: 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.

For 4: 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.

For 5: 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