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Westkost [7]
1 year ago
9

Hi pls help pls pls. this is due at 11:59 help

Chemistry
1 answer:
Sveta_85 [38]1 year ago
7 0

Answer:

mass of salt 6.72g

Explanation:

mass of the combine mass with salt- mass of the empty beaker

36.176g−29.456g=6.72g

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What did J. J. Thomson observe when he applied electric voltage to a cathode ray tube in his famous experiment?
Nataly_w [17]

The glass opposite to the negative electrode started to glow. Hence, option B is correct.

<h3>What is a cathode ray tube?</h3>

A cathode-ray tube (CRT) is a specialized vacuum tube in which images are produced when an electron beam strikes a phosphorescent surface.

J.J. Thomson, through his famous Cathode ray experiment, proved that all atoms contain small negatively charged particles known as electrons. In the experiment, he applied electric voltage across a cathode ray tube. a fluorescent material coating was done on the positive side. When the voltage was applied, the positive side has glowing dots.

Hence, option B is correct.

Learn more about the cathode ray tube here:

brainly.com/question/14409449

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6 0
2 years ago
2c4H10 +13O2 —&gt;8CO2+10H2O to find how many moles of oxygen would react with 4.3 mol C4H10
Elena-2011 [213]

Answer:

\large\boxed{\text{28 mol of O$_{2}$}}

Explanation:

             2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

n/mol:      4.3  

13 mol of O₂ react with 2 mol of 2C₄H₁₀

\text{Moles of O}_{2} = \text{4.3 mol C$_{4}$H$_{10}$} \times \dfrac{\text{13 mol O}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \textbf{28 mol O}_{2}\\\\\text{The reaction requires $\large\boxed{\textbf{28 mol of O$_{2}$}}$}

5 0
3 years ago
Is MgCO3 organic or inorganic
schepotkina [342]
Its inorganic as MgCO3 is contains no carbon more hydrogen which is a crutial component of all organic compounds 
4 0
3 years ago
Compute 6.28×1013+7.30×1011.
Simora [160]
6.28×1013+7.30×1011 this =13741.94
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Rus_ich [418]
The angle- angle similarity postulate. It states that if a triangle has two equal corresponding angles the angles are similar.
6 0
3 years ago
Read 2 more answers
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