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cluponka [151]
3 years ago
9

If 2.0 mL of 6.0M HCl is used to make a 500.0-mL aqueous solution, what is the molarity?

Chemistry
2 answers:
RideAnS [48]3 years ago
5 0

Answer:

0.024M

Explanation:

Data obtained from the question include:

C1 = 6M

V1 = 2mL

C2 =?

V2 = 500mL

The molarity of the diluted solution can be obtained as follows:

C1V1 = C2V2

6 x 2 = C2 x 500

Divide both side by 500

C2 = (6 x 2) /500

C2 = 0.024M

The molarity of the diluted solution is 0.024M

pentagon [3]3 years ago
4 0

Answer:

The molarity is 0.024 mol / L

Explanation:

Molarity = Number of moles / Volume of solution

= (6.0 mol/L x 2.0 x 10^{-3} L) / 0.500 L

= 0.024 mol / L

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100gm of a 55% (M/M) nitric acid solution is to be diluted to 20% (M/M) nitric acid.
yawa3891 [41]

Volume of H₂O added = 175 ml

<h3>Further explanation</h3>

Given

100 gm of a 55% (M/M)  and 20% (M/M) nitric acid solution

Required

waters added

Solution

starting solution

mass H₂O = 45%=45 g

%mass of H₂O in new solution = 100%-20%=80%

Can be formulated for %mass H₂O :

\tt \dfrac{45+x}{100+x}=80\%\\\\45+x=0.8(100+x)\\\\45+x=80+0.8x\\\\35=0.2x\rightarrow x=175~g

For water mass=volume(density = 1 g/ml)

So volume added = 175 ml

7 0
3 years ago
C6H12O6 + 6O2 + 38ADP + 38Pi =&gt; 6CO2 + 6H2O + 38ATP is the chemical equation for
Jobisdone [24]

Answer:

C6H12O6 + 6O2 + 38ADP + 38Pi => 6CO2 + 6H2O + 38ATP is the chemical equation for cellular respiration

Explanation:

Cellular respiration is the process by which cells breakdown glucose molecules to produce energy in the form of ATP molecules and release waste products such as carbon dioxide and water molecules. Cellular respiration involves a series of reaction pathways such as glycolysis, pyruvate oxidation, citric acid cycle and the oxidative phosphorylation pathway.

The first step of glycolysis breaks down a glucose molecule to release two pyruvate molecules.

In pyruvate oxidation, two molecules of pyruvate are oxidized to acetyl-CoA molecules.

In the citric acid cycle, the acetyl-CoA molecules are used to produce the electron carriers NADH and FADH2.

In the oxidative phosphorylation pathway, NADH and FADH2 donate their electrons to oxygen and ATP molecules are produced using the energy of electron transfer and proton-pumping.

The overall equation for cellular respiration is given as:

C6H12O6 + 6O2 + 38ADP + 38Pi => 6CO2 + 6H2O + 38ATP

3 0
3 years ago
What is the orbital hybridization of a central atom that has one lone pair and bonds to:
nirvana33 [79]

sp^3  is the orbital hybridization of a central atom that has one lone pair and bonds to three other atoms.

<h3>What is orbital hybridization?</h3>

In the context of valence bond theory, orbital hybridization (or hybridisation) refers to the idea of combining atomic orbitals to create new hybrid orbitals (with energies, forms, etc., distinct from the component atomic orbitals) suited for the pairing of electrons to form chemical bonds.

For instance, the valence-shell s orbital joins with three valence-shell p orbitals to generate four equivalent sp3 mixes that are arranged in a tetrahedral configuration around the carbon atom to connect to four distinct atoms.

Hybrid orbitals are symmetrically arranged in space and are helpful in the explanation of molecular geometry and atomic bonding characteristics. Usually, atomic orbitals with similar energies are combined to form hybrid orbitals.

Learn more about Hybridization

brainly.com/question/22765530

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5 0
2 years ago
Ion R2+ có phân lớp ngoài cùng là 3p6
amm1812
Ok and bro???? Like what even lol
4 0
3 years ago
PLEASE HELP ME!!!
kondor19780726 [428]
<h3>Answer:</h3>

0.424 J/g °C

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Thermochemistry</u>

Specific Heat Formula: q = mcΔT

  • q is heat (in Joules)
  • m is mass (in grams)
  • c is specific heat (in J/g °C)
  • ΔT is change in temperature
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] m = 38.8 g

[Given] q = 181 J

[Given] ΔT = 36.0 °C - 25.0 °C = 11.0 °C

[Solve] c

<u>Step 2: Solve for Specific Heat</u>

  1. Substitute in variables [Specific Heat Formula]:                                             181 J = (38.8 g)c(11.0 °C)
  2. Multiply:                                                                                                             181 J = (426.8 g °C)c
  3. [Division Property of Equality] Isolate <em>c</em>:                                                         0.424086 J/g °C = c
  4. Rewrite:                                                                                                             c = 0.424086 J/g °C

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.424086 J/g °C ≈ 0.424 J/g °C

6 0
3 years ago
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