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lisabon 2012 [21]
3 years ago
7

2KClO3 —> 2KCl + 302 this equation represents both a ___ reaction AND a(n) ___ reaction.

Chemistry
1 answer:
andriy [413]3 years ago
6 0

Answer:

option D--Decomposition; oxidation-reduction

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In the following reaction, 451.4g of lead reacts with excess oxygen forming 356.7g of lead (II) oxide. Calculate the percent yie
Sholpan [36]
Given the reaction 2Pb(s)+O2(g)= 2PbO(s) and a reactant amount of 451.4 grams, we are asked for the yield of the reaction. The amount of lead present produces 451.4/207.2 *( 2/2) *(223.2) via 100% conversion, 486.26 grams lead (II) oxide. hence the percent yield is 356.7g /<span>486.26 g or equal to 73.35 percent</span>
8 0
3 years ago
What are the values for % Cl and % F, respectively, for Sample II?
horsena [70]

Answer:

Explanation:

The mass percentages of chlorine and fluorine:

  Given:

Mass of chlorine = 5.753g

 Mass of fluorine = 9.248g

Find the total of the masses given:

   Total mass = mass of chlorine + mass of fluorine = 5.753 + 9.248 = 15.001g

% of Cl = \frac{mass of chlorine}{total mass} x 100

% of Cl = \frac{5.753}{15.001} x 100 = 38.35%

% of F =  \frac{mass of fluorine}{total mass} x 100

% of F =  \frac{9.248}{15.001} x 100 = 61.65%

6 0
4 years ago
ryan is a chemistry student that enjoys hot tea. he wants to determine how much ice is needed to cool 250.0 ml of tea to an opti
mestny [16]

Ryan calculates that he requires 4 ice cubes for 250mL of hot tea to reach the optimal drinking temperature by using the specific heat capacity of tea.

A substance's potential to hold heat is indicated by its specific heat capacity. This substance size reflects the amount of heat required to raise a certain volume of a material's temperature by one Kelvin. Specific heat capacity is a distinguishing feature of every substance and is useful for material identification.

Given:

Final temperature of system is 57.8℃

dT1 = 79.1 - 57.8 = 21.3℃

dT2 = 0 - (-8.33) = 8.33℃

dT3 = 57.8 - 0 = 57.8℃

Mass of tea, m1 = 250.0mL = 250.0g = 0.250kg

Specific heat capacity of tea, C1 = 4186 J/kg℃ = Specific heat capacity of water

Specific heat capacity of ice, Ci = 2090 J/kg℃

Mass of each ice cubes, m of i = 18.8g

Latent heat of fusion of ice, Lf = 334 kJ/kg

To find:

No. of ice cubes required = ?

Calculations:

Suppose equilibrium temperature is T, then

Heat released by Tea = Heat gained by Ice

Q1 = Q2

m1x C1x dT1 = mi x Ci x dT2 + mi x Lf + mi x Cw x dT3

0.250x 4186 x 21.3 = mi x (2090 x 8.33 + 3.34 x 10^4 + 4186 x 57.8)

mi = 0.250 x 4186 x 21.3 / (2090 x 8.33 + 3.34 x 10^4 + 4186 x 57.8)

mi = 0.0761kg = 76.1g

Mass of ice required = 76.1g

Number of ice cube required will be:

n = Total mass of ice/mass of each ice cube

n = 76.1/18.8

n = 4.04 ice cubes = 4.0 ice cubes

Result:

Ryan requires 4 ice cubes to bring 250mL of hot tea to the optimal drinking temperature.

Learn more about Specific heat capacity here:

brainly.com/question/24265493

#SPJ4

7 0
2 years ago
4. Mendeleev created the first periodic table by arranging elements in order of
PtichkaEL [24]

Answer:

c

Explanation:

C is the answer for increasing Atomic number

3 0
3 years ago
Which phrase describes one type of freshwater wetland
ikadub [295]

Answer:

Small streams that flow into larger streams and wetlands

Explanation:

Hope this helps, have a great day (;

6 0
3 years ago
Read 2 more answers
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