John Fang has 5 apples, Jonny ate 3, how many are left!

How many liters of carbon dioxide will be produced when 89.5 L of ethane are burned? (One mole of any gas occupies 22.4 L under

yan [13]

Answer:

179 L of CO2

Explanation:

Given the equation of the reaction;

C2H6(g) + 7/2 O2(g) -------> 2CO2(g) + 3H2O(g)

Now 1 mole of ethane yields 2 moles of CO2 from the balanced reaction equation

1 mole of a gas occupies 22.4 L volume so,

22.4 L of ethane yields 44.8 L of CO2

89.5 L of ethane yields 89.5 * 44.8/22.4 = 179 L of CO2

4 0

2 years ago

Which statement describes an Arrhenius acid

liq [111]

Answer:

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs. Hence, the correct statement is arrhenius acid produces hydrogen ions in solution.

3 0

3 years ago

Which of these statements best compares the pair of line segments with the vertex?

liraira [26]

The line segment is describe as a line which joints two points. The word segment is denotes the end points of the line. The symbol of the line segment is AB. This symbol is indicating a line from the point A to the point B.

Generally, the line segment has two endpoints. The meeting point of two line segments is called as the vertex and also it is considered as a common endpoint on the other hand line is a straight path of points that goes on and on in two directions
Vertex is a point where two rays meet OR where the sides of a polygon meet OR the point where three or more edges of a solid figure meet and the angle is Formed by two rays that have the same endpoint

To know more about line segment visit :

brainly.com/question/25727583

#SPJ9

3 0

1 year ago

Match the following aqueous solutions with the appropriate letter from the column on the right.1. 0.19 m AgNO3 2. 0.17 m CrSO4 3

vichka [17]

Answer:

0.13 m of $Mn(NO_3)_2$ → Highest boiling point

0.19 m of $AgNO_3$ → Second Highest boiling point

0.17 m of $CrSO_4$ → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

Explanation:

Elevation in boiling is given by :

$\Delta T_b=i\times k_b\times m$

Where :

i = van't Hoff factor

$k_b$ = Molal Elevation constant of solvent

m = molaity of the solution

1) 0.19 m of $AgNO_3$

$AgNO_3\rightarrow Ag^++NO_3^{-}$

i = 2 (electrolyte)

Molality of the solution = 0.19

Elevation is boiling point of solution:

$\Delta T_b=2\times k_b\times 0.19 m$

$\Delta T_b=0.38 m\times k_b$

2) 0.17 m of $CrSO_4$

$CrSO_4\rightarrow Cr^{2+}+SO_4^{2-}$

i = 2 (electrolyte)

Molality of the solution = 0.17

Elevation is boiling point solution :

$\Delta T_b=2\times k_b\times 0.17 m$

$\Delta T_b=0.34 m\times k_b$

3) 0.13 m of $Mn(NO_3)_2$

$Mn(NO_3)_2\rightarrow Mn^{2+}+2NO_3^{-}$

i = 3 (electrolyte)

Molality of the solution = 0.13

Elevation is boiling point solution :

$\Delta T_b=3\times k_b\times 0.13 m$

$\Delta T_b=0.39 m\times k_b$

4) 0.31 m Sucrose (nonelectrolyte)

i = 1 ( non electrolyte)

Molality of the solution = 0.31 m

Elevation is boiling point solution :

$\Delta T_b=1\times k_b\times 0.31 m$

$\Delta T_b=0.31 m\times k_b$

Higher the value of elevation in temperature higher will be the boiling point of the solution .

The decreasing order of solution from highest boiling point to lowest boiling point is :

$0.39 m\times k_b>0.38 m\times k_b>0.34 m\times k_b>0.31 m\times k_b$

0.13 m of $Mn(NO_3)_2$ → Highest boiling point

0.19 m of $AgNO_3$ → Second Highest boiling point

0.17 m of $CrSO_4$ → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

6 0

3 years ago

Write a balanced half-reaction for the oxidation of aqueous hydrazine N2H4 to gaseous nitrogen N2 in basic aqueous solution

Zigmanuir [339]

Answer:

N2H2(aq) + 2OH^-(aq) ----------> N2(g) + 2H2O(l) + 2e

Explanation:

Hydrazine is mostly used in thermal engineering as an anticorrosive agent. Hydrazine can be oxidized in aqueous solution as shown in the equation above. Oxidation has to do with loss of electrons and increase in oxidation number.

The oxidation number of nitrogen in the equation increased from -1 in hydrazine on the lefthand side of the reaction equation to zero in nitrogen on the right hand side of the reaction equation. Two electrons were lost in the process as shown.

7 0

2 years ago