Which particle has a mass of 9.11 x 10^-28g and charge of -1?
A. electron
B. proton
C. neutron​

Answer:

35g/l

Explanation:

Salinity is practically the saltiness of the water, in basic terms. It's basically the amount of salt dissolved in water. Ocean water has a salinity of around 35g/l (that's about 3.5% of which is dissolved salt in water!). However, Atlantic Ocean (the ocean with the saltiest water), can range up to 37g/l.

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Answer:

i dont know mate

Explanation:

Answer:

a. True

Explanation:

There is strong inhibition of Carbon Anhydrase by Aceta-zolamide Carbonic Anhydrase. The drug acetazolamide is used as diuretic which increase the urine production in human body. It lowers pressure in eye in glaucoma.

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: 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]

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:

[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:

Mass=Moles × RFM

Mass= 0.28M× 164

Mass= 45.92 grammes

Explanation:

elements are carbons and hydrogen

Answer:

Carbon and Hydrogen.

Explanation:

It’s in the name Hydro (H) Carbon (C)

Answer:

123456788012346778901234567890

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.

b) (√)

c)(✓)

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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:

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

Answer:

a. 599 cm³

b. 1.49 g/cm³

Explanation:

Volume is the amount of space an object occupies. Since this is a brick, the object is a rectangular prism. The formula for the volume of a rectangular prism is the product of length, width, and height.

[tex]V= l *w*h[/tex]

The brick's length (l) is 13.77 centimeters, the width (w) is 8.50 centimeters, and the height (h) is 5.12 centimeters. Substitute these values into the formula.

[tex]V= 13.77 \ cm * 8.50 \ cm * 5.12 \ cm[/tex]

Multiply the numbers together.

[tex]V= 117.045 \ cm^ 2* 5.12 \ cm[/tex]

[tex]V= 599.2704 \ cm^3[/tex]

The original measurements have at least 3 significant figures, so our answer must have 3. For the number we calculated, that is the ones place. The 2 in the tenths place tells us to leave the 9 in the ones place.

[tex]V \approx 599 \ cm^3[/tex]

[tex]\bold {The \ volume \ of \ the \ brick \ is \ approximately \ 599 \ cubic \ centimeters}}[/tex]

Density is the amount of matter in a specified space. The formula for density is mass over volume.

[tex]d= \frac{m}{v}[/tex]

The mass of the brick is 895.3 grams and we just found the volume to be 599.2704 cubic centimeters. Substitute the values into the formula.

[tex]d= \frac{895.3 \ g}{599 \ cm^3}[/tex]

Divide.

[tex]d= 1.494657763 \ g/cm^3[/tex]

Round to three significant figures. For the number we calculated, that is the hundredth place. The 4 in the thousandth place tells us to leave the 9 in the hundredth place.

[tex]d \approx 1.49 \ g/cm^3[/tex]

[tex]\bold {The \ density\ of \ the \ brick \ is \ approximately \ 1.49 \ grams /cubic \ centimeters}}[/tex]

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:

i. HCIO2 - 2.9

ii. CH3COOH - 2.64

iii. HF - 3.27

iv. HNO2 - 2.67

Explanation:

The Ph value differ for base and acid. A neutral solution will have Ph value of 7. For acid the Ph value is less than 7 and for base Ph value is greater than 7. The highest Ph value for the given acids is 3.27 for hydroflouric acids.

The pH of the resulting solution is 3.16. Acetic acid having the formula [tex]CH_3COOH[/tex] would produce the highest pH at the equivalence point in a weak acid-strong base titration.

Kingsley adds 49.28 mL of NaOH to 250.00 mL of the HCOOH solution.

The neutralization reaction resulted in 0.098 moles of HCOOH and 0.025 moles of HCOO⁻ left in the solution.

We have to determine the pH of the resulting solution and identify the acid that would produce the highest pH at the equivalence point in a weak acid-strong base titration.

pH, a quantitative measure of the acidity or basicity of aqueous or other liquid solutions.

Kingsley adds 49.28 mL of NaOH to 250.00 mL of the HCOOH solution.

Total volume of the solution = 49.28 mL + 250.0 mL = 299.28 mL = 0.29928 L

At neutralization point, HCOOH = 0.098 mol = [tex]\frac{0.098\ mol}{0.29928\ L} = 0.32745\ M[/tex]

So, Concentration of HCOOH at neutralization point = 0.32745 M

And [tex]HCOO^-[/tex] at neutralization point = 0.025 mol = [tex]\frac{0.025\ mol}{0.29928\ L} = 0.08353\ M[/tex]

So, Concentration of HCOO⁻ at neutralization point = 0.08353 M

Now, [tex]K_a[/tex] value of HCOOH = [tex]1.77\times10^{-4}[/tex] (reference value}

We know that [tex]pK_a = -logK_a[/tex]

[tex]pK_a = -log(1.77\times10^{-4}) = 3.75[/tex]

Now, HCOOH  is acid and HCOO⁻ is the conjugate base.

According to Henderson Hasselblach equation:

[tex]pH = pK_a + log\frac{[conjugate\ base]}{[acid]}\\ \therefore pH = 3.75 + log\frac{[HCOO^-]}{[HCOOH]}\\pH = 3.75 + log(\frac{0.08353}{0.32745})\\pH = 3.75 +(-0.59)\\pH = 3.16[/tex]

Therefore, the pH of the resulting solution is 3.16.

When weak acid is titrated with a strong base, the pH at the neutralization point is always greater than 7. It is because the conjugate base produced by weak acid in neutralization reaction is comparatively stronger in nature which produces [tex]OH^-[/tex] at neutralization point after reacting with water. so the solution is basic. As a result, pH is more than 7.

Now, the weaker is the acid, the stronger is the conjugate base, and vice-versa. Therefore, the concentration of [tex]OH^-[/tex] ions is more at the neutralization point. So more will be the pH value.

Given acids with the corresponding [tex]K_a[/tex] value are shown below:

[tex]K_a(HF) =6.8\times10^{-4}\\K_a(HNO_2) =4.5\times10^{-4}\\K_a(HClO_2) =1.1\times10^{-2}\\K_a(CH_3COOH) =1.8\times10^{-5}[/tex]

We know that lower the value of [tex]K_a[/tex] of  acid, the weaker is the acid and vice-versa. Hence, [tex]CH_3COOH[/tex] has lowest [tex]K_a[/tex] value compared to other acids. Therefore, [tex]CH_3COOH[/tex]  would  produce the highest pH at the equivalence point in a weak acid-strong base titration.

Hence, The pH of the resulting solution is 3.16. Acetic acid having the formula [tex]CH_3COOH[/tex] would produce the highest pH at the equivalence point in a weak acid-strong base titration.

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Answer:.......

Explanation:

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]

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:

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:

62.5 ml of 0.032 M  KMnO₄ are required to react with 50.0 ml of 0.100 molar H₂C₂O₄ in the presence of excess H₂SO₄

Explanation:

The balanced reaction is:

2 KMnO₄ + 5 H₂C₂O₄ + 3 H₂SO₄ → K₂SO₄ + 2 MnSO₄ + 8 H₂O + 10 CO₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Molarity or Molar Concentration is the number of moles of solute that are dissolved in a certain volume.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]

Molarity is expressed in units [tex]\frac{moles}{liter}[/tex]

In this case, 50 mL (0.05 L) of 0.1 M H₂C₂O₄ react. So, replacing the data in the definition of molarity:

[tex]0.1 M=\frac{number of moles of solute}{0.05 L}[/tex]

Solving:

number of moles of solute= 0.1 M*0.05 L

number of moles of solute= 0.005 moles

So, 0.005 moles of H₂C₂O₄ react.  Then you can apply the following rule of three: if by stoichiometry 5 moles of H₂C₂O₄ react with 2 moles of KMnO₄, 0.005 moles of H₂C₂O₄ react with how many moles of KMnO₄?

[tex]moles of KMnO_{4} =\frac{0.005moles of H_{2} C_{2} O_{4}* 2moles of KMnO_{4} }{5moles of H_{2} C_{2} O_{4} }[/tex]

moles of KMnO₄= 0.002 moles

Knowing that the molarity of KMnO₄ is 0.032 M, replacing in its definition and solving:

[tex]0.032 M=\frac{0.002 moles}{volume}[/tex]

[tex]volume=\frac{0.002 moles}{0.032 M}[/tex]

volume= 0.0625 L= 62.5 mL

62.5 ml of 0.032 M  KMnO₄ are required to react with 50.0 ml of 0.100 molar H₂C₂O₄ in the presence of excess H₂SO₄

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 ≈

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:

2.5 × 10² L

Explanation:

Step 1: Given and required data

Step 2: Convert 25 °C to Kelvin

We will use the following expression.

K = °C + 273.15 = 25 + 273.15 = 298 K

Step 3: Calculate the volume (V) of the balloon

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T / P

V = 9.7 mol × (0.0821 atm.L/mol.K) × 298 K / 0.955 atm = 2.5 × 10² L

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:

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)