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charle [14.2K]
2 years ago
6

Indicate whether each of the following statements about Joseph Priestley’s experiment is an observation or an inference.

Chemistry
2 answers:
irina [24]2 years ago
5 0

Answer:

1) Inf.

2) Observation

3) Observation

Genrish500 [490]2 years ago
4 0

Answer:

Inference  — The candle stayed lit longer because the plant gave off oxygen.

Observation — The candle went out when it was placed in a closed jar.

Observation — The candle stayed lit longer in closed jar when a plant was present.

Explanation:

Tap the THANKS button and RATE ⭐️

(If helpful)

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What is the ratio of the diffusion rates of cl2 and o2? rate cl2 : o2 = 0.47 0.67 0.45 0.69 1.5?
matrenka [14]
The  ratio  of  the diffusion  rate  of Cl2  and   O2   is 1.5

    calculation
rate  of diffusion Cl2/ rate   of  diffusion  O2 =√  molar mass of Cl2/ molar mass  of O2

molar  mass of Cl2  = 35.5 x2 = 71  g/mol
molar  mass O2  = 16 x2 =32g/mol

that  is    rate  of diffusion Cl2/ rate  of  diffusion of O2  =√ 71/ 32=  1.5
7 0
2 years ago
Describe a single measurement that you could make that shows that a kettle is less than 100% efficient (providing evidence that
alexgriva [62]

Answer:

As you haven't explained what measurements you took before solving this problem, I will explain the general procedure to evaluate the efficiency of a kettle. I hope it helps you. I´ll send an attachement file with the full answer, since I couldn't write it here.

I assume that the material that is going to be heated in the kettle is water.

1- You have to boil water in it and take the time it takes to its boiling point (in seconds).  

2- You have to evaluate the amount of energy the water absorbed Q with the efficiency formula which I explain in the attachement file.

3- Divide Q by the time it took to bring the water to boiling so you can have the power it consumed.

4- You divide the last value you obtained by the Kettles's power rating.  

5- Multiply the last value by 100 to obtain a percentage value of efficiency.

Explanation:

Efficiency is the ration of a machine's useful work, in this case how much energy the water absorbed to get to its boiling point divided by the time it took to get to this point, and the total energy expended, in this case the kettles's power rating.

7 0
2 years ago
The number of sp2 hybrid orbitals on the carbon atom in CO32– is
Vladimir79 [104]
The number of sp2 hybrid orbitals on the carbon atom in CO32– is 3. Because hybrids = combination of 2 different types of orbitals
sp2 = 1/3 s character + 2/3 p character
7 0
2 years ago
Read 2 more answers
The average density of a carbon-fiber-epoxy composite is 1.615 g/cm3. the density of the epoxy resin is 1.21 g/cm3 and that of t
ra1l [238]
The composite material is composed of carbon fiber and epoxy resins. Now, density is an intensive unit. So, to approach this problem, let's assume there is 1 gram of composite material. Thus, mass carbon + mass epoxy = 1 g.

Volume of composite material = 1 g / 1.615 g/cm³ = 0.619 cm³
Volume of carbon fibers = x g / 1.74 g/cm³ 
Volume of epoxy resin = (1 - x) g / 1.21 g/cm³ 

a.) V of composite = V of carbon fibers + V of epoxy resin
0.619 = x/1.74 + (1-x)/1.21
Solve for x,
x = 0.824 g carbon fibers
1-x = 0.176 g epoxy resins

Vol % of carbon fibers = [(0.824/1.74) ÷ 0.619]*100 =<em> 76.5%</em>

b.) Weight % of epoxy = 0.176 g epoxy/1 g composite * 100 = <em>17.6%</em>
     Weight % of carbon fibers = 0.824 g carbon/1 g composite * 100 = <em>82.4%</em>
4 0
2 years ago
Copper(II) sulfide, CuS, is used in the development of aniline black dye in textile printing. What is the maximum mass of CuS wh
Naya [18.7K]

Answer:

1.82 g   is the maximum mass of CuS.

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

<u>For CuCl_2 : </u>

Molarity = 0.500 M

Volume = 38.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 38.0×10⁻³ L

Thus, moles of CuCl_2 :

Moles=0.500 \times {38.0\times 10^{-3}}\ moles

<u>Moles of CuCl_2  = 0.019 moles </u>

<u>For (NH_4)_2S : </u>

Molarity = 0.600 M

Volume = 42.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 42.0×10⁻³ L

Thus, moles of (NH_4)_2S :

Moles=0.600 \times {42.0\times 10^{-3}}\ moles

<u>Moles of (NH_4)_2S  = 0.0252 moles </u>

According to the given reaction:

CuCl_2_{(aq)}+(NH_4)_2S_{(aq)}\rightarrow CuS_{(s)}+2NH_4Cl_{(aq)}

1 mole of CuCl_2 reacts with 1 mole of (NH_4)_2S

So,  

0.019 mole of CuCl_2 reacts with 0.019 mole of (NH_4)_2S

Moles of (NH_4)_2S = 0.019 mole

Available moles of (NH_4)_2S = 0.0252 mole

<u>Limiting reagent is the one which is present in small amount. Thus, CuCl_2 is limiting reagent.</u>

The formation of the product is governed by the limiting reagent. So,

1 mole of CuCl_2 gives 1 mole of CuS

0.019 mole of CuCl_2 gives 0.019 mole of CuS

Moles of CuS formed = 0.019 moles

Molar mass of CuS = 95.611 g/mol

Mass of CuS = Moles × Molar mass = 0.019 × 95.611 g = 1.82 g

<u>1.82 g   is the maximum mass of CuS.</u>

5 0
2 years ago
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