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tester [92]
3 years ago
5

Gases Test Chemistry

Chemistry
1 answer:
Soloha48 [4]3 years ago
4 0

Answer:

6.64 l is the answer.......

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Sooeme and I just wanted you guys know how much I appreciate you guys wanna offer me and I have your
Marat540 [252]

Answer:

Your answers are solids, liquids, and gases.

Explanation:

These are the three states of matter.

<em>Please</em><em> like</em><em> and</em><em> mark</em><em> brainliest</em><em>!</em>

<em>Hope</em><em> it</em><em> helps</em><em>!</em>

4 0
3 years ago
Read 2 more answers
How many moles are there in 24.00 g of NaCl
Elis [28]

Answer:

The answer to your question is 0.41 moles

Explanation:

Data

moles of NaCl = ?

mass of NaCl = 24 g

Process

To solve this problem just calculate the molar mass of NaCl, and remember that the molar mass of any substance equals to 1 mol.

1.- Calculate the molar mass

NaCl = 23 + 35.5 = 58.5 g

2.- Use proportions and cross multiplication

               58.5 g of NaCl ------------------- 1 mol

               24.0 g               ------------------- x

                     x = (24 x 1) / 58.5

                     x = 0.41 moles

6 0
3 years ago
Which of the following reactions could be used to extract the lead from the lead nitrate?
alexgriva [62]

Answer:

Option b. Decomposition

Followed by a reduction process using charcoal

Explanation:

Lead can be obtained from lead nitrate by thermal decomposition of lead nitrate as shown below:

2Pb(NO3)2 —> 2PbO + 4NO2 + O2

The PbO obtained is reduced by charcoal(C) to obtain the metallic Pb as shown below:

2PbO + C —> Pb + CO2

4 0
3 years ago
What biological processes can add co2 to the atmosphere?
RSB [31]
Cellar respiration simple if u need explanation just search up
6 0
3 years ago
a solution with a transmittance of 0.44 is analyzed in a spectrophotometer with 6% stray light. calculate the absorbance reporte
IgorLugansk [536]

The absorbance reported by the defective instrument was 0.3933.

Absorbance A = - log₁₀ T

Tm = transmittance measured by spectrophotometer

Tm = 0.44

Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654

True absorbance can be calculated by true transmittance, Tm = T+S(α-T)

S = fraction of stray light = 6%= 6/100 = 0.06

α= 1, ideal case

T = true transmittance of the sample

Tm = T+S(α-T)

now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233

therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)

i.e; 0.3933

To know more about transmittance click here:

brainly.com/question/17088180

#SPJ4

3 0
1 year ago
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