Is hot tea a substance or a mixture

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The coefficients, in order of the blanks, are 5, 2, 5, 2, 4.

Answer:

When you move the burette slider to the top of a flask and add about 25 mL of NaOH to the flask, you will cause a concentration of OH- molecules. This will make the solution become a basic solution and make the litmus paper blue.

Explanation:

After reading your question, we can see that you are carrying out a test to discover the nature of the pH of a solution. This type of test uses litmus paper, which is an indicator of the presence of acids and bases, being able to determine the pH of a solution. This paper is soaked in organic ink and when placed in an acidic solution, it is red in color. However, when placed in a basic solution it has a blue color.

An acidic solution is one that has a high concentration of H+ atoms and has the ability to donate electrons. The basic solution, on the other hand, has a high concentration of OH- and has the capacity to receive electrons.

When you move the burette slider to the top of a flask and add about 25 mL of NaOH to the flask, you will cause a concentration of OH- molecules. This will make the solution become a basic solution and make the litmus paper blue, that is, the solution has the basic pH.

Answer:

0.047 moles

Explanation:

From the question,

Applying

PV = nRT............. Equation 1

Where P = Pressure of CH₄ gas, V = Volume of CH₄ gas, n = number of moles of CH₄ gas, R = Molar gas constant.

make n the subject of the equation

n = PV/RT............. Equation 2

Given: P = 660 torr = (660×0.00131579) atm = 0.868 atm, V = 1300 mL = (1300/1000) L = 1.3 L, T = 20 °C = (20+273) K = 293 K

Constant: R = 0.082 atm.dm³/K.mol

Substitute these values into equation 2

n = (0.868×1.3)/(0.082×293)

n = 1.1284/24.026

n = 0.047 mole

Answer:

As blood travels through the body, oxygen is used up, and the blood becomes oxygen poor. Oxygen-poor blood returns from the body to the heart through the superior vena cava (SVC) and inferior vena cava (IVC), the two main veins that bring blood back to the heart.

Explanation:

Answer:Redox and combustion.

Explanation:

C3H8) reacts with oxygen in the air, so  C3H8(g) + 5O2(g) = 3CO2(g) + 4H2O(g) Classification: Redox and combustion.

Answer: precipitate

Explanation:

Answer: It is a basic solution with pH =8.51

Explanation:

We were given that hydroxide ion concentration [OH-] as 3.26 x 10-6 M

But We know thar

[OH-] [H+] = 1 × 10^-14

To get the Hydrogen ion [H+] concentration, we have that

[H+] = 1 × 10^-14 M/[OH-]

= 1 × 10^-14 M/3.26 x 10-6 M

    =  3.067 x 10^-9 M

But, pH = - log [H+]

Therefore,

pH = - log (3.067 x 10^-9)

pH=8.51

when  pH > 7, The solution is basic, therefore a solution with pH =8.51 is basic.

Answer: Answer is A it will decrease soil erosion

Explanation:it just will trust me

Answer:

A.

Explanation:

the soil will not loosen when u till it

Freeze-Thaw Weathering

This action can widen the cracks in the rock, and when the temperature rises above freezing, the ice thaws, allowing the water to seep further into the cracks. As this process of freezing and thawing happens repeatedly, the rock begins to weaken and eventually breaks apart into angular fragments.

Answer:

Physical weathering is caused by the effects of changing temperature on rocks, causing the rock to break apart. The process is sometimes assisted by water. ... Freeze-thaw occurs when water continually seeps into cracks, freezes and expands, eventually breaking the rock apart.

Explanation:

hope this helps

Answer: The volume of [tex]NaOH[/tex] required is 25.0 ml

Explanation:

According to the neutralization law,

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1[/tex] = basicity [tex]HNO_3[/tex] = 1

[tex]M_1[/tex] = molarity of [tex]HNO_3[/tex] solution = 2.00 M

[tex]V_1[/tex] = volume of  [tex]HNO_3[/tex] solution = 50.0 ml

[tex]n_2[/tex] = acidity of [tex]NaOH[/tex] = 1

[tex]M_1[/tex] = molarity of [tex]NaOH[/tex] solution = 4.00 M

[tex]V_1[/tex] = volume of  [tex]NaOH[/tex] solution =  ?

Putting in the values we get:

[tex]1\times 2.00\times 50.0=1\times 4.00\times V_2[/tex]

[tex]V_2=25.0ml[/tex]

Therefore, volume of [tex]NaOH[/tex] required is 25.0 ml

The volume of 4.00 M NaOH is required to exactly neutralize 50.0 ml of a solution of a 2.00 M solution of [tex]HNO_3[/tex] is 25.0 ml.

[tex]HNO_3[/tex] is nitric acid also known as the spirit of niter. It is colorless in its original form but appears yellow-colored.

By the neutralization law

[tex]n_1M_1V_1 = n_2M_2V_2[/tex]

n1 = 1

M1 = 2.00

M2 = 4.00

V1 = 50.0 ml

n2 = NaOH = 1

Putting the value in the formula

[tex]1 \times 2.00 \times 5.00 ml = 1 \times 4.00 \times V_2\\\\V_2 = 25.0\; ml[/tex]

Thus, the volume of NaOH required is 25.0 ml.

Learn more about nitric acid

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Answer:

The cell wall surrounds the plasma membrane of plant cells and provides tensile strength and protection against mechanical and osmotic stress. It also allows cells to develop turgor pressure, which is the pressure of the cell contents against the cell wall.

Explanation:

ion

Answer:

hi

Explanation:

Light is the result of electrons moving between defined energy levels in an atom, called shells.

Answer:

Explanation:

3Pb(ii)(NO3)2 + 2Fe(iii) ( C2H3O2 )3   ===> 3Pb( C2H3O2)2 + 2Fe(NO3)3

I know you have likely been given this as a homework assignment, but in real life I doubt you would be able to see any reaction taking place. The products are likely both soluble and both ionize, so I don't think you will get any kind of reaction taking place.

The equation however is balanced.

As the blades rotate, they experience frictional forces as they `rub' against the air; this knocks electrons around, causing the blades to build up a net charge. The charged dust particles then stick to the charged areas of the blades. "The leading edge of the blades usually develops the thickest layer of dust.

To solve this we must be knowing each and every concept related to dust. Therefore,  frictional forces allow the ceiling fan´s blade to collect dust so quickly.

Dust stirred up by automobiles on the road may account for 33% overall air pollution. Deposits of vehicle or commercial exhaust gas, particles through tire and braking wear, dust from paved streets or potholes, plus dust from building sites all contribute to road dust.

As the blades revolve, frictional forces cause them to 'rub' against the air, knocking electrons around and allowing the blades to accumulate a net charge. The electrified dust particles are subsequently attracted to the charged portions of the blades. Typically, the sharp end of the blades creates the heaviest coating of dust.

Therefore,  frictional forces allow the ceiling fan´s blade to collect dust so quickly.

To know more about dust, here:

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Answer:

Bonds between atoms are broken/ or new bonds are formed

Explanation:

Due to chemical changes.

Answer:

1.14 M

Explanation:

First divide the mass of NaOH by the molar mass of NaOH. Then divide by the volume to get molarity.

34.2 g / 39.997 g/mol = 0.855 mol NaOH

0.855 mol NaOH / 0.750 L = 1.14 M NaOH

Answer:

what did you end up guessing

Explanation:

Answer:

Cell lol

Explanation:

answer B

"C" is correct.

The Law of Conservation of Mass states that matter can be neither created nor destroyed. This is evident in chemical equations as you will always start and end with the same amount of atoms of every element. They are all still there; they have just been rearranged.

Answer: C) The number of atoms of each element is the same on each side of the equation.

Explanation: Mass cannot be created nor destroyed.

Based on the rate of disappearance of HI, the rate of appearance of I₂ is 0.225 mol/L/s.

The rate of a reaction is the rate at which reactant molecules are converted to product or the rate at which product molecules are formed.

From the equation of the reaction, 2 moles of HI produces 1 mole of I₂.

The rate of disappearance of HI is -0.45 mol/L/s

The rate of appearance of I₂ will be 0.45/2 mol/L/s = 0.225 mol/L/s.

Therefore, the rate of appearance of I₂ is 0.225 mol/L/s.

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Answer:

670.45 K

Explanation:

Given that,

Initial volume, V₁ = 1.22 L

Initial temperature, T₁ = 286 K

Final volume, V₂ =  2.86 L

We need to find the new temperature of the gas. The relation between the volume and the temperature is given by :

[tex]\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\T_2=\dfrac{V_2T_1}{V_1}\\\\T_2=\dfrac{2.86\times 286}{1.22}\\\\T_2=670.45\ K[/tex]

So, the new temperature of the gas is equal to 670.45 K.

Answer:

D and R are in the same period

Explanation:

remember groups are vertical and periods are horizontal!

Answer:

The correct answer is - Frequency is the number of wavelengths, which is measured in hertz.

Explanation:

Frequency is the number of waves that go through a fixed point at a particular time. Hertz is the SI unit for frequency which means that one hertz is equal to a unit number of waver passes in a unit time to a fixed point.

As the frequency of a wave increases which means the number of waves increases in the unit time, the shorter the wavelength will be.

a higher frequency wave has more energy than a lower frequency wave with the same amplitude.

Answer:

3.25 M

Explanation:

Number of moles = mass/ molar mass

mass of  NaOH = 65 g

Molar mass of NaOH = 40 g/mol

Number of moles of NaOH = 65g/40 g/mol = 1.625 moles

Let us recall that;

n = CV

n = number of moles of solute

C = concentration of solution

V = volume of solution

C = n/V

n = 1.625 moles

V = 500/1000 L = 0.5 L

C = 1.625 moles/0.5

C = 3.25 M

Answer:

Reduced air pressure also has a serious effect on the human body. Reduced air pressure means that there is less oxygen available to the body. For lungs to inflate, the air pressure in your lungs has to be less than the air outside the lungs. This is because air moves from high-pressure areas to low-pressure areas.

Explanation:

Reduced air pressure also has a serious effect on the human body. Reduced air pressure means that there is less oxygen available to the body. For lungs to inflate, the air pressure in your lungs has to be less than the air outside the lungs. This is because air moves from high-pressure areas to low-pressure areas.