what is irrigation?Describe two methods of irrigation which conserve water​

Answer:

0.030 M

Explanation:

Step 1: Make an ICE chart

        2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

I        0.060      0.050          0

C         -2x            -x            +2x

E     0.060-2x  0.050-x       2x

Step 2: Find the value of x

The concentration of SO₃ at equilibrium is 0.040 mol/L. Then,

2x = 0.040

x = 0.020

Step 3: Calculate the concentration at equilibrium of O₂

[O₂] = 0.050 - x = 0.050 - 0.020 = 0.030 M

The equilibrium concentration of oxygen is 0.030 M.

A reversible reaction is a reaction that can move either in the forward or in the reverse direction. We have the reaction; 2SO2(g) + O2(g) ⇄ 2SO3(g). We can now set up the ICE table as shown below;

  2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

I        0.060      0.050          0

C         -2x            -x            +2x

E     0.060-2x  0.050-x       2x

At equilibrium;

2x = 0.040 M

x = 0.040 M/2 =  0.020 M

For oxygen;

0.050-x  

0.050 M - 0.020 M = 0.030 M

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Answer:

the number of protons in a atom is unique to each element

Explanation:

protons are about 99.86% as massive neutrons. The number of protons in a atom is unique to each element .For example carbon atoms have six protons in an atom is referred to as the atomic number of that element

Answer:

0.105 M

Explanation:

First we convert 25.3 grams of sucrose into moles, using sucrose's molar mass:

Now we calculate the molarity of the solution, using the given volume and the calculated number of moles:

Answer:

0.93 M

Step-by-step Explanation:

First, we have to calculate the molar mass (MM) of ZnSO₄ by using the molar mass of each chemical element:

MM(ZnSO₄) = 65.4 g/mol Zn + 32 g/mol S + (16 g/mol x 4) = 161.4 g/mol

Then, we divide the mass of ZnSO₄ into its molar mass to obtain the number of moles:

moles ZnSO₄ = mass/MM = 150 g/(161.4 g/mol)= 0.93 mol

Since the molarity of a solution expresses the number of moles of solute per liter of solution, we calculate the molarity (M) as follows:

M = moles ZnSO₄/1 L =  0.93 mol/1 L = 0.93 M

Answer:

a. 0.7119 g

Explanation:

To solve this question we need to know that all carbon of the compound will react producing CO2 and all Hydrogen producing H2O.

Thus, we can find the mass of C and the mass of H and by difference regard to the 2.584g of the compound we can find the mass of oxygen as follows:

Moles CO2 = Moles C -Molar mass: 44.01g/mol-

5.874g CO2 * (1mol/44.01g) = 0.1335 moles CO2 = 0.1335 moles C

Mass C -Molar mass: 12.01g/mol-:

0.1335 moles C * (12.01g /mol) = 1.6030g C

Moles H2O -Molar mass: 18.01g/mol-

2.404gH2O * (1mol / 18.01g) = 0.1335 moles H2O * (2mol H / 1mol H2O) = 0.267 moles H

Mass H -Molar mass: 1g/mol-

0.267 moles H * (1g/mol) = 0.2670g H

Mass Oxygen =

Mass O = 2.584g compound - 1.6030g C - 0.2670g H

Mass O = 0.714g O ≈

Solution :

Spontaneous Process

A spontaneous process is defined as the process that occurs without the help of any external aid or inputs. A spontaneous process is a natural process which occurs naturally in the environment.

Non Spontaneous process

A non spontaneous process is a process which does not occur naturally. Some inputs are provided for the process to occur. Energy from external source is applied into the process to start the process.

The following processes are :

1. Melling of ice   ---- Spontaneous

2 Rusting of iron  --- Spontaneous

3. Marble going down the spiral.   --- spontaneous

4. Going up the hill  ---- Non spontaneous

5. Keeping the food fresh from spoilage​  --- Non spontaneous

Answer:

0.21 g

Explanation:

The equation of the reaction is;

NaCl(aq) + AgNO3(aq) -----> NaNO3(aq) + AgCl(s)

Number of moles of NaCl= 0.0860 g /58.5 g/mol = 0.00147 moles

Number of moles of AgNO3 = 30/1000 L × 0.050 M = 0.0015 moles

Since the reaction is 1:1, NaCl is the limiting reactant.

1 mole of NaCl yields 1 mole of AgCl

0.00147 moles of NaCl yields 0.00147 moles of AgCl

Mass of precipitate formed = 0.00147 moles of AgCl × 143.32 g/mol

= 0.21 g

Answer:

The correct equation is "[tex]\frac{[Ni(H_2O)_3 (NH_3)]^{2+}}{[Ni(H_2O)_4]^{2+} [NH_3]}[/tex]".

Explanation:

According to the question,

Throughout an aqueous solution, [tex]Ni^{2+}[/tex] exist as [tex][Ni(H_2O)_4]^{2+}[/tex]

So,

⇒ [tex][Ni(H_2O)_4]^{2+} + 4NH_3 \rightleftharpoons [Ni(NH_3)_4]^{2+} + H_2O[/tex]

⇒ [tex]K_f = \frac{[Ni(NH_3)_4]^{2+}}{[Ni(H_2O)_4^{2+}] [NH_3]^4}[/tex]

Here, we have excluded [tex][H_2O][/tex] as concentration of water will be const.

Now,

This formation of [tex][Ni(NH_3)_4]^{2+}[/tex] proceeds via several steps,

Step 1:

⇒ [tex][Ni(H_2O)_4]^{2+}+NH_3 \rightleftharpoons [Ni(H_2O)_3 (NH_3)]^{2+} + H_2O[/tex]

⇒ [tex]K_1 = \frac{[Ni(H_2O)_3 (NH_3)]^{2+}}{[Ni(H_2O)_4]^{2+} [NH_3]}[/tex]

Answer:

NON-METAL

Explanation:

Phosphorus is a non-metal that sits just below nitrogen in group 15 of the periodic table. This element exists in several forms, of which white and red are the best known.

Answer:

i dont know mate

Explanation:

Answer:.......

Explanation:

Ribose is a reducing sugar. A reducing sugar is a carbohydrate that can undergo a redox reaction, in which it donates electrons to another chemical species.

This is usually observed when the sugar opens its ring structure to form an aldehyde or ketone functional group.

Ribose, a five-carbon sugar, can form an open-chain structure with an aldehyde functional group. In this form, it can donate electrons and act as a reducing agent in certain chemical reactions, such as the reduction of other compounds like Benedict's reagent during laboratory tests for reducing sugars.

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Answer:

The pressure increases by a factor of four.

Explanation:

Let's consider a gas at a given temperature and pressure (T₁, P₁). The absolute temperature of a gas is increased four times (T₂ = 4 T₁) while maintaining a constant volume. We can assess the effect on the pressure (P₂) by using Gay Lussac's law.

P₁/T₁ = P₂/T₂

P₂ = P₁ × T₂/T₁

P₂ = P₁ × 4 T₁/T₁

P₂ = 4 P₁

The pressure increases by a factor of four.

Answer:

Different materials will change temperature at different rates when exposed to the same amount of thermal energy. This is because each substance has its own specific heat.

Explanation:

Answer:

[tex]\boxed {\boxed {\sf 8.10 \ L}}[/tex]

Explanation:

This question asks us find the volume of a gas sample given a change in temperature. Since the pressure remains constant, we only are concerned with the variables of temperature and volume.

We will use Charles's Law. This states the volume of a gas is directly proportional to the temperature of a gas. The formula is:

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

The gas starts at a volume of 2.70 liters and a temperature of 25.0 degrees Celsius.

[tex]\frac {2.70 \ L}{25.0 \textdegree C}=\frac{V_2}{T_2}[/tex]

The temperature is increased to 75.0 degrees Celsius, but the volume is unknown.

[tex]\frac {2.70 \ L}{25.0 \textdegree C}=\frac{V_2}{75.0 \textdegree C}[/tex]

We are solving for the volume at 75 degrees Celsius, so we must isolate the variable V₂.

It is being divided by 75.0 °C. The inverse operation of division is multiplication, so we multiply both sides of the equation by 75.0 °C.

[tex]75.0 \textdegree C *\frac {2.70 \ L}{25.0 \textdegree C}=\frac{V_2}{75.0 \textdegree C} * 75.0 \textdegree C[/tex]

[tex]75.0 \textdegree C *\frac {2.70 \ L}{25.0 \textdegree C}= V_2[/tex]

The units of degrees Celsius (° C) cancel.

[tex]75.0 *\frac {2.70 \ L}{25.0}= V_2[/tex]

[tex]75.0 *0.108 \ L = V_2[/tex]

[tex]8.1 \ L = V_2[/tex]

The original measurements have 3 significant figures, so our answer must have the same. Currently, the answer has 2. If we add another 0, the value of the answer does not change, but the number of sig figs does.

[tex]8.10 \ L = V_2[/tex]

The volume of this gas sample at 75.0 degrees Celsius is 8.10 Liters.

Answer:

Temperature affects the kinetic energy in a gas the most, followed by a comparable liquid, and then a comparable solid. The higher the temperature, the higher the average kinetic energy, but the magnitude of this difference depends on the amount of motion intrinsically present within these phases.

Explanation:

Liquids have more kinetic energy than solids. When a substance increases in temperature, heat is being added, and its particles are gaining kinetic energy. Because of their close proximity to one another, liquid and solid particles experience intermolecular forces. These forces keep particles close together.

Answer:

56.9 mmoles of acetate are required in this buffer

Explanation:

To solve this, we can think in the Henderson Hasselbach equation:

pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])

To make the buffer we know:

CH₃COOH  +  H₂O  ⇄   CH₃COO⁻  +  H₃O⁺     Ka

We know that Ka from acetic acid is: 1.8×10⁻⁵

pKa = - log Ka

pKa = 4.74

We replace data:

5.5 = 4.74 + log ([acetate] / 10 mmol)

5.5 - 4.74 = log ([acetate] / 10 mmol)

0.755 = log ([acetate] / 10 mmol)

10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)

5.69 = ([acetate] / 10 mmol)

5.69 . 10 = [acetate] → 56.9 mmoles

Answer:

2 H₂O(l) + 2 e⁻ ⇒ H₂(g) + 2 OH⁻(aq)

Explanation:

Let's consider the unbalanced half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution.

H₂O(l) ⇒ H₂(g)

First, we will perform the mass balance. We will balance oxygen atoms by multiplying H₂O by 2 and adding 2 OH⁻ to the right side.

2 H₂O(l) ⇒ H₂(g) + 2 OH⁻(aq)

Then, we perform the charge balance by adding 2 electrons to the left side.

2 H₂O(l) + 2 e⁻ ⇒ H₂(g) + 2 OH⁻(aq)

Answer:

Partial pressure of H₂ = 499 mmHg

Partial pressure of Ar = 481 mmHg

Explanation:

We'll begin by calculating the number of mole of each gas. This can be obtained as follow:

For Hydrogen:

Molar mass of H₂ = 2 × 1 = 2 g/mol

Mass of H₂ = 0.291 g

Mole of H₂ =?

Mole = mass /molar mass

Mole of H₂ = 0.291/ 2

Mole of H₂ = 0.1455 mole

For Argon:

Molar mass of Ar = 40 g/mol

Mass of Ar = 5.62 g

Mole of Ar =?

Mole = mass /molar mass

Mole of Ar = 5.62 / 40

Mole of Ar = 0.1405 mole

Next, we shall determine the mole fraction of each gas. This can be obtained as follow:

Mole of H₂ = 0.1455 mole

Mole of Ar = 0.1405 mole

Total mole = 0.1455 + 0.1405

Total mole = 0.286 mole

Mole fraction of H₂ (nₕ₂) = mole of H₂ / total mole

Mole fraction of H₂ (nₕ₂) = 0.1455/0.286

Mole fraction of H₂ (nₕ₂) = 0.509

Mole fraction of Ar (nₐᵣ) = mole of Ar / total mole

Mole fraction of Ar (nₐᵣ) = 0.1405/0.286

Mole fraction of Ar (nₐᵣ) = 0.491

Finally, we shall determine the partial pressure of each gas. This can be obtained as follow:

For Hydrogen:

Mole fraction of H₂ (nₕ₂) = 0.509

Total pressure (Pₜ) = 980 mmHg

Partial pressure of H₂ (Pₕ₂) =?

Pₕ₂ = nₕ₂ × Pₜ

Pₕ₂ = 0.509 × 980

Partial pressure of H₂ (Pₕ₂) = 499 mmHg

For Argon:

Partial pressure of H₂ (Pₕ₂) = 499 mmHg

Total pressure (Pₜ) = 980 mmHg

Partial pressure of Ar (Pₐᵣ) =?

Pₜ = Pₕ₂ + Pₐᵣ

980 = 499 + Pₐᵣ

Collect like terms

980 – 499 = Pₐᵣ

481 = Pₐᵣ

Partial pressure of Ar (Pₐᵣ) = 481 mmHg

SUMMARY:

Partial pressure of H₂ (Pₕ₂) = 499 mmHg

Partial pressure of Ar (Pₐᵣ) = 481 mmHg

Answer: Volume would be 196.15 mL if the temperature were changed to [tex]22^{o}C[/tex] and the pressure to 1.25 atmospheres.

Explanation:

Given: [tex]T_{1} = 35^{o}C = (35 + 273) K = 308 K[/tex],     [tex]V_{1}[/tex] = 256 mL,    

[tex]P_{1}[/tex] = 720 torr (1 torr = 0.00131579 atm) = 0.947368 atm

[tex]T_{1} = 22^{o}C = (22 + 273) K = 295 K[/tex],       [tex]P_{2} = 1.25 atm[/tex]  

Formula used to calculate volume is as follows.

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]

Substitute the values into above formula as follows.

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1 atm \times 256 mL}{308 K} = \frac{1.25 atm \times V_{2}}{295 K}\\V_{2} = 196.15 mL[/tex]

Thus, we can conclude that the volume would be 196.15 mL if the temperature were changed to [tex]22^{o}C[/tex] and the pressure to 1.25 atmospheres.

Answer:

1860g.

Explanation:

It is known that the molar mass of C2H6O2 is 62.08 g/mol.,

Now to  solve for the number of moles of solute, one must multiply both

sides by the volume:

moles of solute = (6.00 M)(5.00 L) = 30.0 mol

Notice since the definition of molarity is mol/L, the

product M × L gives mol, a unit of amount.

Use the molar  mass of C3H8O3, one can convert mol to g:

Mass m =30 mol × 62.08 g/mol

m = 1860g.

Hence, there are 1,860 g of C2H6O2 in the specified amount of

engine coolant.

Answer:

a. discrete analyzer

b. 96 well microplate

c. flow injection analysis

d. colorimeter

Explanation:

96 well microplates are instruments designed for sample collection and throughput screening. If an environment lab has collected 2000 samples then 96 well microplate is best suited application. Discrete analyzer is automated chemical analyzer which performs test on samples kept in discrete cells. Flow injection analysis is approach used for chemical analysis. It injects a plug of sample into a flowing carrier stream. Colorimeter is a device which measures absorbance of wavelength of light by a specific solution.

Answer:

A. Concentration of diluted acid = 0.00389 M

B. Concentration of OH- in AgOH solution = 0.00012 M

Explanation:

A. Using the dilution formula: C1V1 = C2V2

Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

From the data provided, C1 = 0.09714 M, V1 = 10.0 mL, V2 = 250.0 mL and C2 = ?

Making C2 subject of the formula above; C2 = C1V1/V2

C2 = 0.09714 M × 10 / 250 = 0.00389 M

B. Equation of the neutralization reaction is given below:

HCl + AgOH ---> AgCl + H₂O

From the equation, 1 mole of acid neutralizes 1 mole of base

Using the titration formula; CaVa/CbVb = na/nb

Where Ca is the concentration of the acid HCl = 0.00389 M

Va is the volume of acid = 7.93 mL

Cb is the concentration of base, AgOH = ?

Vb is volume of base = 250.0 mL

na/nb = mole ratio of acid and base = 1

Making Cb subject of the formula in the equation above; Cb = CaVa/Vb

Cb = 0.00389 M × 7.93 / 250

Cb = 0.00012 M

Answer:

Mass=Moles × RFM

Mass= 0.28M× 164

Mass= 45.92 grammes

Answer:

123456788012346778901234567890

Answer: The boiling point of the solution is [tex]101.02^oC[/tex]

Explanation:

We are given:

3.5 % (by weight) NaCl

This means that 3.5 g of NaCl is present in 100 g of solution

Mass of solvent = Mass of solution - Mass of solute

Mass of solvent (water) = (100 - 3.5) g = 96.5 g

Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.

The expression for the calculation of elevation in boiling point is:

[tex]\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m[/tex]

OR

[tex]\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)

where,

Boiling point of pure solvent (water) = [tex]100^oC[/tex]

Boiling point of solution = ?

i = Vant Hoff factor = 2 (for NaCl)

[tex]K_b[/tex] = Boiling point elevation constant = [tex]0.512^oC/m[/tex]

[tex]m_{solute}[/tex] = Given mass of solute (NaCl) = 3.5 g

[tex]M_{solute}[/tex] = Molar mass of solute (NaCl) = 36.5 g/mol

[tex]w_{solvent}[/tex] = Mass of solvent (water) = 96.5 g

Putting values in equation 1, we get:

[tex]\text{Boiling point of solution}-(100)=2\times 0.512\times \frac{3.5\times 1000}{36.5\times 96.5}\\\\\text{Boiling point of solution}=(1.02+100)^oC\\\\\text{Boiling point of solution}=101.02^oC[/tex]

Hence, the boiling point of the solution is [tex]101.02^oC[/tex]

Explanation:

elements are carbons and hydrogen

Answer:

Carbon and Hydrogen.

Explanation:

It’s in the name Hydro (H) Carbon (C)

Answer:

C

Explanation:

The specific heat capacity=quantity of heat in joule/mass×change in temperature

from this question the quantity of heat is 75250,the mass is 2000 and the change in temperature is 50-30

which is 20

therefore

c=75250/2000×20

c=75250/40000

c=1.88

I hope this helps