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3 years ago

The process of _______ begins in the mouth, where for is ground into small pieces.

Chemistry

2 answers:

krek1111 [17]3 years ago

5 0

Umm... excretion sorry not sure about that one. Hope this helsp

ziro4ka [17]3 years ago

4 0

B) saliva.

excretion is waste in liquid form as in pee which does not begin in the mouth so it's wrong.
waste obviously does not begin in the mouth.
digestion begins in your stomach, not your mouth.
saliva starts in the mouth and then processes on. So saliva is correct

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The volume of a gas is 27.5 mL at 22.0°C and 0.974 atm. What will the volume be at 15.0°C and 0.993 atm? Use Ideal Gas Law (PV =

Scorpion4ik [409]

Answer:

26.3 mL

Explanation:

Step 1:

Obtaining an appropriate gas law from the ideal gas equation.

This is illustrated below:

From the ideal gas equation:

PV = nRT

Divide both side by T

PV/T = nR

At this stage, we'll assume the number of mole (n) to be constant.

Note: R is the gas constant.

PV/T = constant.

We can thus, write the above equation as:

P1V1/T1 = P2V2/T2

The above equation is called the general gas equation.

Step 2:

Data obtained from the question. This includes the following:

Initial volume (V1) = 27.5 mL

Initial temperature (T1) = 22.0°C = 22.0°C + 273 = 295K

Initial pressure (P1) = 0.974 atm.

Final temperature (T2) = 15.0°C = 15.0°C + 273 = 288K

Final pressure (P2) = 0.993 atm

Final volume (V2) =..?

Step 3:

Determination of the final volume of the gas using the general gas equation obtained. This is illustrated below:

P1V1 /T1 = P2V2/T2

0.974 x 27.5/295 = 0.993 x V2/288

Cross multiply to express in linear.

295x0.993xV2 = 0.974x27.5x288

Divide both side by 295 x 0.993

V2 = (0.974x27.5x288)/(295x0.993)

V2 = 26.3 mL

Therefore, the new volume of the gas is 26.3 mL

4 0

3 years ago

Is mendelevium flammable

Alik [6]

No it is not flammable

6 0

2 years ago

How many moles of HCl would react with 37.1 mL of 0.138 M Sr(OH)2

Lena [83]

Answer is: 0.102 moles of HCl would react.

Balanced chemical reaction:

2HCl(aq) + Sr(OH)₂ → SrCl₂(aq) + 2H₂O(l).

V(Sr(OH)₂) = 37.1 mL ÷ 1000 mL/L.

V(Sr(OH)₂) = 0.0371 L; volume of the strontium hydroxide solution.

c(Sr(OH)₂) = 0.138 M; molarity of the strontium hydroxide solution.

n(Sr(OH)₂) = c(Sr(OH)₂) · V(Sr(OH)₂).

n(Sr(OH)₂) = 0.0371 L · 0.138 mol/L.

n(Sr(OH)₂) = 0.0051 mol; amount of the strontium hydroxide.

From balanced chemical reaction: n(Sr(OH)₂) : n(HCl) = 1 : 2.

n(HCl) = 2 · n(Sr(OH)₂).

n(HCl) = 2 · 0.0051 mol.

n(HCl) = 0.0102 mol; amount of the hydrochloric acid.

5 0

3 years ago

High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen. (a) Determine t

Lemur [1.5K]

Answer:

Concentration of Flourine = 24.756%

Explanation:

Given that :

High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen.

the objective is to determine he concentration of Flourine (in wt%) that must be added if this substitution occurs for 12% of all the original hydrogen atoms.

At standard conditions , the atomic weight of the these compounds are as follows:

Carbon = 12.01 g/mol

Chlorine = 35.45 g/mol

Fluorine = 19.00 g/mol

Hydrogen = 1.008 g/mol

Oxygen = 16.00 g/mol

The chemical formula for polyethylene = (CH₂ - CH₂)ₙ

Therefore, for two carbons, there will be 4 hydrogens;

i.e

(CH₂ - CH₂)₂

( C₂H₄ - C₂H₄ )

Suppose the number of original hydrogen = 4moles

number of moles of Flourine F = 12% of 4

= 0.12 × 4

= 0.48 mol

∴ the number of remaining moles of Hydrogen is:

= 4 - 0.48

= 3.52 moles

number of moles of Carbon = 2 moles

∴ the mass of flourine F = number of moles of F × molar mass of F

= 0.48 × 19

= 9.12

The total mass of the compound now is = (0.48 × 19 ) + (3.52 × 1) + (2× 12)

= 9.12 + 3.52 + 24

= 36.64

Concentration of Flourine = (mass of flourine/total mass) × 100

Concentration of Flourine = (9.12/36.84 ) × 100

Concentration of Flourine = 0.24756 × 100

Concentration of Flourine = 24.756%

3 0

3 years ago

What is the mass of solute in a 500 mL solution of 0.200 M

Fofino [41]

16.4 grams is the mass of solute in a 500 mL solution of 0.200 M .

sodium phosphate

Explanation:

Given data about sodium phosphate

atomic mass of Na3PO4 = 164 grams/mole

volume of the solution = 500 ml or 0.5 litres

molarity of sodium phosphate solution = 0.200 M

The formula for molarity will be used here to know the mass dissolved in the given volume of the solution:

The formula is

molarity = $\frac{number of moles of solute}{volume in litres}$

putting the values in the equation, we get

molarity x volume = number of moles

0.200 X 0.5= number of moles

number of moles = 0.1 moles

Atomic mass x number of moles = mass

putting the values in the above equation

164 x 0.1 = 16.4 grams

16.4 grams of sodium phosphate is present in 0.5 L of the solution to make a 0.2 M solution.

8 0

3 years ago