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Alecsey [184]
3 years ago
13

What sequence should be followed when conducting a laboratory investigation? Make observations, gather experimental data, form a

conclusion, state a problem Define a problem, form a hypothesis, gather experimental data, form a conclusion. Form a hypothesis, form a conclusion, gather experimental data, define a problem Gather experimental data, make observations, form a conclusion, for a hypothesis.​
Chemistry
2 answers:
MArishka [77]3 years ago
5 0

Answer:

Make observations, gather experimental data, form a conclusion, state a problem Define a problem, form a hypothesis, gather experimental data, form a conclusion.

Explanation:

AURORKA [14]3 years ago
3 0

Answer:

Define a problem, form a hypothesis, gather experimental data, form a conclusion

You might be interested in
A sample of table sugar (sucrose, C12H22O11) has a mass of 1.202 g.
Archy [21]

Answer:

a) 0.003512 moles

b) Moles C= 0.04214 moles carbon

Moles H = 0.07726 moles hydrogen

Moles O = 0.03863 moles of oxygen

c) C atoms = 2.54 *10^22 carbon atoms

H atoms = 4.65 *10^22 hydrogen atoms

O atoms = 2.33 *10^22 oxygen atom

Explanation:

Step 1: Data given

Mass of sucrose = 1.202 grams

Molar mass of sucrose = 342.3 g/mol

Step 2: Calculate moles of sucrose

Moles sucrose = Mass sucrose / molar mass sucrose

Moles sucrose = 1.202 grams / 342.3 g/mol

Moles sucrose = 0.003512 moles

Step 3: Calculate moles of each element

For 1 mol of C12H22O11 we have 12 moles of carbon, 22 moles of hydrogen and 11 moles of oxygen

Moles C: 12*0.003512 = 0.04214 moles carbon

Moles H: 22* 0.003512 = 0.07726 moles hydrogen

Moles O: 11* 0.003512 = 0.03863 moles of oxygen

Step 4: Calculate the number of atoms

C atoms = 6.022 *10^23 / mol * 0.04214 moles = 2.54 *10^22 atoms carbon

H atoms = 6.022 * 10^23 / mol * 0.07726 moles = 4.65 *10^22 atoms H

O atoms = 6.022 * 10^23 / mol * 0.03863 moles = 2.33 *10^22 atoms O

0.003512 moles of sucrose containse 6.022 *10^23 * 0.003512 = 2.11 * 10^21 sucrose molecules

7 0
3 years ago
A closed system initially containing 1×10^-3 hydrogen 2×10^-3M iodine at 448 degree Celsius and is allowed to reach equilibrium.
GaryK [48]

Answer:

Kc = 50.5

Explanation:

We determine the reaction:

H₂  +  I₂   ⇄   2HI

Initially we have 0.001 molesof H₂

and 0.002 moles of I₂

If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.

           H₂     +      I₂      ⇄   2HI

In:     0.001       0.002           -

R:       x                 x                2x

Eq:  0.001-x    0.002-x      0.00187  

x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted

So in the equilibrium we have:

0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵  moles of H₂

0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂

Expression for Kc is =  (HI)² / (H₂) . (I₂)

0.00187 ² /  6.5×10⁻⁵ . 1.065×10⁻³ = 50.5

5 0
3 years ago
What is the Chemical Formula and Net Ionic Equations for all three solutions.
AlladinOne [14]

Answer:

See answer below

Explanation:

As you are asking for chemical formula and ionic equation, then, I will assume that after the station #3 below, are the solutions you are requiring.

You are also not specifing if you want for example, result of solution 1 + solution 3. If you need that, please post that on another question.

Now for the chemical formula, you need to identify the elements in all 3 solutions, and also the type of compound.

<u>1. Solution 2 Potassium Iodide: </u>

In this case we have Potassium on one side, and Iodine on the other side, the symbol for those are K and l. As Potassium have the +1 oxidation state, cause is the only one that it can have, when it's next to an halide like chlorine or bromine, it will form a binary salt. The halides, usually work with the lowest oxydation state. In the case of Iodide it will be -1, so, the formula will be:

KI

And the net ionic equation will be the chemical equation that shows how the charges and atoms are balanced. In this case it would be:  

K⁺ + I⁻ ------> KI

<u>2. Solution 1 and 3, Lead (II) nitrate and Sodium carbonate: </u>

In this case I will work with both, because both of the solution are tertiary compounds. In this cases, we have two tertiary salt, The Sodium symbol is Na, and is working with it oxydation state +1. Carbonate is an anion and have the formula CO₃ working with the oxydation state -2. Lead can work with oxidation state +2 and +4. It's symbol is Pb. Nitrate is NO₃ and works with oxydation state -1 instead.

The chemical formula and ionic equation for both will be:

Lead(II) nitrate: Pb(NO₃)₂

Sodium Carbonate: Na₂CO₃

And the net equations:

Lead nitrate: Pb²⁺ + 2NO₃⁻ ------> Pb(NO₃)₂

Copper sulfate: 2Na⁺ + CO₃²⁻ -------> Na₂CO₃

Hope this helps

6 0
3 years ago
What is the formula for photo synthesis
pshichka [43]

6CO2 + 6H20 + (energy) → C6H12O6 + 6O2

aka

Carbon dioxide+water+energy(from light)=glucose and oxygen

6 0
3 years ago
Read 2 more answers
Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c
Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>

<em />

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>

<em />

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M

6 0
3 years ago
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