Cal is titrating 57.7 mL of 0.311 M HBr with 0.304 M Ba(OH)2. How many mL of Ba(OH)2 does Cal need to add to reach the equivalence point?

Explanation:

The required concentration of [tex]HNO_3[/tex] M1 =0.222 M.

The required volume of [tex]HNO_3[/tex] is V1 =225 mL.

The standard solution of [tex]HNO_3[/tex] is M2 =16 M.

The volume of standard solution required can be calculated as shown below:

Since the number of moles of solute does not change on dilution.

The number of moles [tex]n=molarity * volume[/tex]

[tex]M_1.V_1=M_2.V_2[/tex]

[tex]V2=\frac{M_1.V_1}{M_2} \\=0.222M x 225 mL / 16 M\\=3.12 mL[/tex]

Hence, 3.12 mL of 16 m nitric acid is required to prepare 0.222 M and 225 mL of nitric acid.

Answer:

B. 0.0000005461m

I used the method of moving the decimal.

Answer:

Ethyn gas ( acetylene gas )

Explanation:

All group II carbides react with water to form ethyn gas apart from beryllium which produces methane gas.

[tex]{ \sf{CaC_{2(s)} + 2H _{2} O_{(l)} → Ca(OH) _{2(s)} + C _{2} H _{2(g)} }}[/tex]

Answer:

D. releases heat energy.​

Explanation:

Answer:

Explanation:  Độ thẩm thấu của NaCl 0.9% và glucose 5% lần lượt là 308 và 278 ... Dung dịch natri clorid sử dụng trong pha thuốc tiêm truyền thường dùng

Answer: 1090°C

Explanation: According to combined gas laws

(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2

where P1 = initial pressure of gas = 80.0 kPa

V1 = initial volume of gas = 10.0 L

T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K

P2 = final pressure of gas = 107 kPa

V2 = final volume of gas = 20.0 L

T2 = final temperature of gas

Substituting the values,

(80.0 kPa × 10.0 L) ÷ (513 K) = (107 kPa × 20.0 L) ÷ T2

T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)

T2 = 513 K × (1.3375) × (2)

T2 = 1372.275 K

T2 = (1372.275 - 273) °C

T2 = 1099 °C

1090 degree Celsius

hope it helps

Answer

A

Explanation:

due to high specific heat capacity it loses heat and has low temperature

Answer:

hydrogen nitrate + sodium hydrochlorate- sodium nitrate+ water + co2 (acid base reaction)

silver nitrate + calcium chloride - silver chloride+ calcium nitrate ( double displacement reaction)

hydrogen + nitrogen - ammonia gas ( simple contact reaction)

hydrogen peroxide - water + oxygen ( single displacement reaction)

Hope it helps :)

Answer:

NaCl, Na⁺,Cl⁻.

MgCl₂, Mg²⁺, Cl⁻.

CaO, Ca²⁺, O²⁻.

Li₃P, Li⁺, P³⁻.

Al₂S₃, Al³⁺, S²⁻.

Ca₃N₂, Ca²⁺, N³⁻.

FeCl₃, Fe³⁺, Cl⁻.

FeO, Fe²⁺, O²⁻.

Cu₂S, Cu⁺, S²⁻.

Cu₃N₂, Cu²⁺, N³⁻.

ZnO, Zn²⁺, O²⁻.

Ag₂S, Ag⁺, S²⁻.

K₂CO₃, K⁺, CO₃²⁻.

NaNO₃, Na⁺, NO₃⁻.

Ca(HCO₃)₂, Ca²⁺, HCO₃⁻.

Al(OH)₃, Al³⁺,OH⁻.

Li₃PO₄, Li⁺, PO₄³⁻.

K₂SO₄, K⁺, SO₄²⁻.

Explanation:

Sodium chloride. NaCl, formed by the cation Na⁺ and the anion Cl⁻.

Magnesium chloride. MgCl₂, formed by the cation Mg²⁺ and the anion Cl⁻.

Calcium oxide. CaO, formed by the cation Ca²⁺ and the anion O²⁻.

Lithium phosphide. Li₃P, formed by the cation Li⁺ and the anion P³⁻.

Aluminum sulfide. Al₂S₃, formed by the cation Al³⁺ and the anion S²⁻.

Calcium nitride. Ca₃N₂, formed by the cation Ca²⁺ and the anion N³⁻.

Iron(III)chloride. FeCl₃, formed by the cation Fe³⁺ and the anion Cl⁻.

Iron(II)oxide. FeO, formed by the cation Fe²⁺ and the anion O²⁻.

Copper(I)sulfide. Cu₂S, formed by the cation Cu⁺ and the anion S²⁻.

Copper(II)nitride. Cu₃N₂, formed by the cation Cu²⁺ and the anion N³⁻.

Zinc oxide. ZnO, formed by the cation Zn²⁺ and the anion O²⁻.

Silver sulfide. Ag₂S, formed by the cation Ag⁺ and the anion S²⁻.

Potassium carbonate. K₂CO₃, formed by the cation K⁺ and the anion CO₃²⁻.

Sodium nitrate. NaNO₃, formed by the cation Na⁺ and the anion NO₃⁻.

Calcium bicarbonate. Ca(HCO₃)₂, formed by the cation Ca²⁺ and the anion HCO₃⁻.

Aluminum hydroxide. Al(OH)₃, formed by the cation Al³⁺ and the anion OH⁻.

Lithium phosphate. Li₃PO₄, formed by the cation Li⁺ and the anion PO₄³⁻.

Potassium sulfate. K₂SO₄, formed by the cation K⁺ and the anion SO₄²⁻.

Answer:

The various uses of water :

1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.

2. Water is used as a universal solvent.

3. water maintains the temperature of our body.

4. Water helps in digestion in our body.

5 .water is used in factories and industries.

6. Water is used to grow plants , vegetables and crops.

Answer:

The various use of water are;

I) Cooking.

ii) Drinking

III) Bathing

iv) Generating hydro- electricity

v) Construction work etc

Answer:

a. 1.12 L

Explanation:

Step 1: Write the balanced equation for the photosynthesis

6 CO₂(g) + 6 H₂O(l) ⇒ C₆H₁₂O₆(s) + 6 O₂(g)

Step 2: Calculate the moles corresponding to 2.20 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

2.20 g × 1 mol/44.01 g = 0.0500 mol

Step 3: Calculate the moles of O₂ produced

The molar ratio of CO₂ to O₂ is 6:6. The moles of O₂ produced are 6/6 × 0.0500 mol = 0.0500 mol

Step 4: Calculate the volume occupied by 0.0500 moles of O₂ at STP

At STP, 1 mole of O₂ occupies 22.4 L.

0.0500 mol × 22.4 L/1 mol = 1.12 L

Answer:

Mg+F2= Mgf2

Explanation:

F 2 is an oxidizing agent, Mg is a reducing agent. ; Pale-yellow to greenish gas with a pungent, irritating odor.

Answer:

an alcohol is a Hydroxyl group due to the OH~ that is associated with it's molecules

The functional group found in an alcohol is Hydroxyl . Therefore, the correct option is option D.

A functional group in organic chemistry is a substituent and moiety inside a molecule that triggers the molecule's distinctive chemical processes. No matter how the rest of a molecule is made up, the very same functional group would experience the same or a similar set of chemical events.

This permits the design of synthetic chemistry as well as the methodical forecasting of chemical reactions as well as the behaviour of chemical molecules. Other functional groups close by can affect a functional group's reactivity. Retrosynthetic analysis can be used to design organic synthesis by using functional group interconversion. The functional group found in an alcohol is Hydroxyl .

Therefore, the correct option is option D.

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Answer:

d

Explanation:

Answer:

The correct option is (c)

Answer:

the law of conservation of mass

The compound that has been soluble in water will be LiCl. Thus option E is correct.

The compounds that form stable bonds in water and are easy to dissociate are soluble in water. Generally, ionic compounds are easily soluble in water.

It has been an ionic compound, but the dissociation of AgCl will simultaneously form AgCl again making the compound insoluble in water.

It has been a covalent compound, thus it has been insoluble in water.

The compound is covalent in nature, thus it has been hard to dissociate in water.

The copper in the compound has been hard to dissolve and precipitate in the solution.

It has been an ionic compound and can be easily dissociated in the water.

Thus the compound that has been soluble in water will be LiCl. Thus option E is correct.

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Answer:

vinegr and soda ..................

........

Answer:

acid + base = salt

so the answer is vinegar and soda

Explanation:

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Answer:

See the answer below

Explanation:

From the original equation in the image, the mole ratio of C:CO2:CO is 1:1:2. This means that for every 1 mole of C and CO2, 2 moles of CO would be produced.

Now, looking at the simulation below the equation of the reaction, 3 moles of C and 8 moles of CO2 were supplied as input. Applying this to the original equation of reaction, C seems to be a limiting reagent for the reaction because the ratio of C to CO2 should 1:1.

Hence, taking all the 3 moles of C available means that only 3 moles out of the available 8 for CO2 would be needed. 3 moles c and 3 moles CO2 means that 6 moles CO would be produced (remember that the ratio remains 1:1:3 for C, CO2, and CO). This means that 5 moles CO2 would be leftover.

In other words, all the 3 moles C would be consumed, 3 out of 8 moles CO2 would be consumed, and 6 moles CO would be produced while 5 moles CO2 would be leftover.

Answer:

Table salt: answer salt

Tea: answer acidic

Carbonated drinks: answer acidic

Baking powder: answer acid and base

Detergent: answer acid and basic

Alum: answer acidic

Explanation:

I hope this helps. Enjoy your day!

[tex]\boxed{\sf {AlCl_3\atop Aluminium\:Chloride}+{H_2\atop Hydrogen}\longrightarrow {Al\atop Aluminium}+{HCl\atop Hydrochloric\:acid}}[/tex]

Balanced Equation:-

[tex]\boxed{\sf {2AlCl_3\atop Aluminium\:Chloride}+{3H_2\atop Hydrogen}\longrightarrow {2Al\atop Aluminium}+{6HCl\atop Hydrochloric\:acid}}[/tex]

Answer:

"[tex]6.7\times 10^{-4} \ atm[/tex]" is the right answer.

Explanation:

Given:

Partial pressure of [tex]N_2[/tex],

= 0.20 atm

Partial pressure of [tex]H_2[/tex],

= 0.15 atm

[tex]K_p = 1.5\times 10^3[/tex] at [tex]400^{\circ} C[/tex]

As we know,

⇒ [tex]K_p = \frac{pN_2\times pH_2^3}{pNH_3^2}[/tex]

By putting the values, we get

    [tex]1.5\times 10^3=\frac{0.20\times (0.15)^3}{pNH_3^2}[/tex]

        [tex]pNH_3^2 = \frac{0.000675}{1.5\times 10^3}[/tex]

                    [tex]=6.7\times 10^{-4} \ atm[/tex]

Answer:

An alkaline is a substance that dissolves in water to produce hydroxyl ions (OH-)

Explanation:

The pH range of an alkaline is from 8–14.

Acidic pH ranges from 0–6.9.

Answer

Iridium(IV)Nitrite

The correct IUPAC name of the Ir(NO₂)₄ compound is Iridium(IV)Nitrite.

Whether it's in a continuous chain or just a ring, the largest chain of carbons joined by a single bond serves as the basis for IUPAC nomenclature.

A chemical compound would seem to be a substance that contains numerous similar molecules made of atoms from different elements joined by chemical bonds.

The given compound is Ir(NO₂)₄. It can be seen that 4 nitro group is attached with Ir and its coordination number is 4. Hence, the IUPAC name will be Iridium(IV)Nitrite.

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Answer:

pH = 12.43

Explanation:

...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it.

To solve this question we need to know that hidrazoic acid reacts with KOH as follows:

HN3 + KOH → KN3 + H2O

Moles KOH:

0.5716L * (0.2900mol /L) =0.1658 moles of KOH

Moles HN3:

0.2127L * (0.6800mol/L) = 0.1446 moles HN3

As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:

0.1658 moles - 0.1446 moles =

0.0212 mol KOH

In 212.7mL + 571.6mL = 784.3mL = 0.7843L

The molarity of KOH = [OH-] is:

0.0212 mol KOH / 0.7843L = 0.027M = [OH-]

The pOH is defined as -log [OH-]

pOH = -log 0.027M

pOH = 1.57

pH = 14 - pOH

pH = 12.43

Answer:

In metrology, measurement uncertainty is the expression of the statistical dispersion of the values attributed to a measured quantity.By international agreement, this uncertainty has a probabilistic basis and reflects incomplete knowledge of the quantity value. It is a non-negative parameter.

Answer:

See explanation

Explanation:

For a substance to dissolve in another, there must be some sort of interaction between the substances.

Recall that like dissolves like. That is, polar substances dissolve polar substances and non polar substances dissolve nonpolar substances.

Hydrocarbons are nonpolar hence they do not dissolve in polar sulphuric acid. Methyl benzoate is polar hence it dissolve in polar sulphuric acid.

The equation showing the ions is depicted in the image attached to this answer.

Chemistry 11 Solutions

978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 8 Solutions and Their Properties • MHR | 85

Amount in moles, n, of the NaCl(s):

NaCl

2.5 g

m

n

M

58.44 g

2

4.2778 10 m l

ol

o

/m

u

Molar concentration, c, of the NaCl(aq):

–2 4.2778 × 10 mol

0.100

0.42778 mol/L

0.43 mol

L

/L

n

c

V

The molar concentration of the saline solution is 0.43 mol/L.

Check Your Solution

The units are correct and the answer correctly shows two significant digits. The

dilution of the original concentrated solution is correct and the change to mol/L

seems reasonable.

Section 8.4 Preparing Solutions in the Laboratory

Solutions for Practice Problems

Student Edition page 386

51. Practice Problem (page 386)

Suppose that you are given a stock solution of 1.50 mol/L ammonium sulfate,

(NH4)2SO4(aq).

What volume of the stock solution do you need to use to prepare each of the

following solutions?

a. 50.0 mL of 1.00 mol/L (NH4)2SO4(aq)

b. 2 × 102 mL of 0.800 mol/L (NH4)2SO4(aq)

c. 250 mL of 0.300 mol/L NH4

+

(aq)

What Is Required?

You need to calculate the initial volume, V1, of (NH4)2SO4(aq) stock solution

needed to prepare each given dilute solution.

The dilution gives the relationship between the molarity and the volume of the solution. The volume of stock solution with a molarity of 1.50 mol/L is 50 mL.

Dilution is said to be the addition of more volume to the concentrated solution to make it less in molar concentration. This tells about the inverse and indirect relationship between the volume and the molar concentration of the solution.

Given,

Initial volume = V₁

Initial molar concentration (M₁) = 1.50 mol/L

Final volume (V₂) = 125 mL = 0.125 L

Final molar concentration (M₂)= 0.60 mol/L

The dilution is calculated as:

M₁V₁ = M₂V₂

V₁ = M₂V₂ ÷ M₁

Substituting the values in the above formula as

V₁ = M₂V₂ ÷ M₁

V₁ = (0.60 mol/L × 0.125 L) ÷ 1.50 mol/ L

V₁ = 0.05 L

= 50 mL

Therefore, 50 mL of stock solution is needed to make a 0.60 mol/L solution.

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Answer:

14720000

Explanation:

1 kg = 1000000 mg

14.72 kg = 14.72 x 1000000

=14720000

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Answer:

B. 1.65 L

Explanation:

Step 1: Write the balanced equation

2 SO₂(g) + O₂(g) ⇒ 2 SO₃(g)

Step 2: Calculate the moles of SO₂

The pressure of the gas is 1.20 atm and the temperature 25 °C (298 K). We can calculate the moles using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.20 atm × 1.50 L / (0.0821 atm.L/mol.K) × 298 K = 0.0736 mol

Step 3: Calculate the moles of SO₃ produced

0.0736 mol SO₂ × 2 mol SO₃/2 mol SO₂ = 0.0736 mol SO₃

Step 4: Calculate the volume occupied by 0.0736 moles of SO₃ at STP

At STP, 1 mole of an ideal gas occupies 22.4 L.

0.0736 mol × 22.4 L/1 mol = 1.65 L

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