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3 years ago

What is the independent variable of Smithers experience?

Chemistry

1 answer:

amm18123 years ago

8 0

Answer:

Smithers thinks that a special juice will increase the productivity of workers. He creates two groups of 50 workers each and assigns each group the same task (in this case, they're ... Group A is given the special juice to drink while they work. Group ... 1. Control Group. Group B. 2. Independent Variable. volume of special juice.

Explanation:

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Melting The change from a _____________ to a ______________.<br><br> Choices: Solid, Liquid, and Gas

Annette [7]

Explanation:

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7 0

3 years ago

Read 2 more answers

How many molecules are in 1.2 moles of H20?

Otrada [13]

<h3>Answer:</h3>

7.226 × 10^23 molecules.

<h3>Explanation:</h3>

A compound is a substance that is made by two or more atoms from different elements.

A mole of a compound contains a number of molecules equivalent to Avogadro's number, 6.022 × 10^23.
That is, one mole of a compound contains 6.022 × 10^23 molecules.

In this case we are given;

Number of moles of H₂O as 1.2 moles

But, 1 mole of H₂O contains 6.022 × 10^23 molecules.

We are required to calculate the number of molecules present;

To calculate the number of molecules we are going to multiply the number of molecules in one mole by the number of moles.

Therefore;

Number of molecules = 1.2 moles × 6.022 × 10^23 molecules/mole

= 7.226 × 10^23 molecules.

Thus, 1.2 moles of water contains 7.226 × 10^23 molecules.

4 0

3 years ago

Reduction of aqueous nitrous acid hno2 to gaseous nitric oxide no in acidic aqueous solution. be sure to add physical state symb

telo118 [61]

Reduction is only one half of the reaction of a redox (reduction-oxidation reaction). It is characterized by the reduction of oxidation number or the gain of electrons. So, you would expect the reaction to have moles of electrons in the reactant side to depict gaining of electrons. The reduction reaction is as follows:

<em>HNO₂ + e⁻ --> NO</em>

Why only 1 e-? Compute the oxidation number of N in the reactant side.
1+x+2(-2) = 0; x = +3
Then, compute the oxidation number of N in the product side.
x -2 = 0; x = +2

So, there is a difference of 1 electron. Hence, 1e-.

5 0

3 years ago

An average adult has a total lung capacity of 6.0 L. How many total grams of air could be held in the lungs at a pressure of 102

satela [25.4K]

You're able to figure this out by using the ideal gas equation PV=nRT Where R=.08 (L*atm/mol*K) Convert 37 C to Kelvin=310 K and 102 kPa to atm=roughly 1 atm Manipulate the ideal gas equation to n=PV/RT Plug in your values (1 atm * 6 L) / (.08 (L*atm/mol*K) * 310 K) All units should cancel except for mol and you should get .24 mol. Multiply the molar mass of air by your answer. .24 mol * 29 g/mol moles should cancel and you should get 6.9 g.

5 0

3 years ago

Read 2 more answers

Consider the chemical equations shown here. C(s) + 2H2(g) → CH4(g) C(s) + 2Cl2(g) → CCl4(g) H2(g) + Cl2(g) → 2HCl(g) What is the

Vikentia [17]

Answer:

The answer to your question is below

Explanation:

(I) C(s) + 2H2(g) → CH4(g)

(II) C(s) + 2Cl2(g) → CCl4(g)

(III) H2(g) + Cl2(g) → 2HCl(g)

Process

Change (I) CH4(g) ⇒ C(s) + 2H2(g)

Add (I) and (II) C(s) + 2Cl2(g) ⇒ CCl4(g)

Result CH4(g) + C(s) + 2Cl2(g) ⇒ CCl4(g) + C(s) + 2H2(g)

Simplify CH4(g) 2Cl2(g) ⇒ CCl4(g) + 2H2(g) (IV)

Multiply (III) by 2 2( H2(g) + Cl2(g) → 2HCl(g))

2 H2(g) + 2Cl2(g) → 4HCl(g) (V)

Add (IV) and (V)

CH4(g) + 2Cl2(g) ⇒ CCl4(g) + 2H2(g)

2 H2(g) + 2Cl2(g) ⇒ 4HCl(g)

CH4(g) + 2Cl2(g) + 2 H2(g) + 2Cl2(g) ⇒ CCl4(g) + 2H2(g) + 4HCl(g)

Simplify

CH4(g) + 4Cl2(g) ⇒ CCl4(g) + 4HCl(g)

4 0

3 years ago