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3 years ago

2. What problem did astronaut Eugene Cernan encounter while on an EVA? How did it relate to

Chemistry

1 answer:

MaRussiya [10]3 years ago

7 0

Answer:

Well he couldn't be pushed back as there isn't a place for him to float back, because of Newton's third law: Every force has an opposite force when Eugene pushes, he will float backwards

Explanation:

Hope this helps!

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When the equation below is balanced, the coefficient of sulfuric acid, h2so4 is: nacn + h2so4 → na2so4 + hcn?

horsena [70]

2NaCN + (1)H2SO4 → Na2SO4 + 2HCN

The coefficient of sulfuric acid is 1.

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3 years ago

How many valence electrons are in an atom of phosphorus

dezoksy [38]

5 Valence electrons .......... Hope it helps, Have a nice day:)

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3 years ago

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Test tubes are used mostly for _____. A. storing experimental samples B. making precise measurements of liquids C. containing ex

Annette [7]

Answer: A) Storing experimental samples

Explanation:

It is a common piece of laboratory glassware that can be made of glass or plastic and is opened at the top and closed at the bottom.

It cannot be used for measurements because there is no graduation indicating the volume.

Althought it can contain extra chemicals left over from an experiment, it is not the main proposal of the glassware that is to store samples.

It cannot be used in a microscope and the object for that is a microscope slide.

8 0

3 years ago

What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2

atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

$Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}$

$Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}$

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

$Mole of AlCl_3 = Mole of Al * Mole ratio$

$Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}$

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = $133.34 \frac{g}{mol}$

Mass of AlCl3 can be calculated using mole formula as:

$1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}$

Multiplying both side by $133.34 \frac{g}{mol}$

$1.32 mole * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34 \frac{g}{mol}$

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

$Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}$

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

$Mole of AlCl_3 = mole of Cl_2 * mole ratio$

$Mole of AlCl_3 = 0.55 mole * \frac{2}{3}$

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

$0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}$

Multiplying both side by

$133.34 \frac{g}{mol}$

$0.367 mol of AlCl_3 * 133.34 \frac{g}{mol} = \frac{mass(g)}{133.34\frac{g}{mol}} * 133.34 \frac{g}{mol}$

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

7 0

3 years ago

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Hellllppppp lons Worksheet

Iteru [2.4K]

The metals will lose electrons while the non metals will gain electrons in order to attain octet structure.

An ion can be cation (positively charged) or anion (negatively charged).

Cations attain octet structure (8) by losing electron(s) while anions become stable or attains octet structure (8) by gaining electron(s).

The remaining elements are completed as follows to attain octet structure;

Element--valence electron--electrons to gain--electrons to lose--ion formed

O ------------ 6 ---------------------- 2 ------------------------ none -------------- $O^{2-}$

Ca -------- 2 ----------------------- none ---------------------- 2 ------------------ $Ca^{2+}$

Br ----------- 7 --------------------- 1 ------------------------ none --------------- $Br^{-}$

S ------------ 6 ----------------------- 2 ------------------------ none --------------- $S^{2-}$

Cl ------------ 7 ----------------------- 1 ------------------------ none ---------------- $Cl^{-}$

K -------------- 1 ----------------------- none ----------------------- 1 ------------------ $K^{+}$

Mg ------------ 2 ---------------------- none ---------------------- 2 ---------------- $Mg^{2+}$

Be ------------- 2 ---------------------- none ---------------------- 2 ---------------- $Be^{2+}$

Learn more here: brainly.com/question/21089350

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3 years ago