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Arte-miy333 [17]
3 years ago
5

The rate of a reaction is measured by how fast a reactant is used up to how fast a product is formed.

Chemistry
2 answers:
Nimfa-mama [501]3 years ago
8 0
Yes it is since rate of reaction is the time taken to complete a reaction
bezimeni [28]3 years ago
5 0
A. True.

Very true. The quicker or slower the reactants are used up the faster or slower the rate of reaction, and the faster or slower the products are formed, the faster or slower the rate of reaction.
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Given the unbalanced equation: ___Al2(SO4)3+___Ca(OH)2—>___Al(OH)3+__CaSO4
nasty-shy [4]

Answer:

The answer to your question is letter B. 9

Explanation:

Unbalanced reaction

                     Al₂(SO₄)₃  +  Ca(OH)₂   ⇒    Al(OH)₃   +   CaSO₄

                     Reactants             Elements      Products

                          2                          Al                     1

                          3                           S                     1

                         14                           O                    7

                          1                           Ca                    1

                          2                           H                     3

Balanced reaction

                    Al₂(SO₄)₃  + 3Ca(OH)₂   ⇒   2Al(OH)₃   +   3CaSO₄

                     Reactants             Elements      Products

                          2                          Al                    2

                          3                           S                    3

                         18                           O                  18

                          3                           Ca                  3

                          6                            H                    6

The sum of the coefficients is 1 + 3+ 2+ 3 = 9

8 0
3 years ago
Calculează masa zaharului si volumul apei necesare pentru prepararea 500g de soluție de zahăr cu partea de masa 20%
Sedbober [7]
T = 20 %  : 20 / 100 = 0.2

m1 = solute 

m2 = Solvent

T = m1 / m1 + m2

0.2 = 500 g / 500 g + m2

0.2 * ( 500 + m2 ) = 500

0.2 * 500 + 0.2 m2 = 500

100 + 0.2 m2 = 500

0.2 m2 = 500 - 100

0.2 m2 = 400

m2 = 400 / 0.2

m2 = 2000 g of water

hope this helps!



7 0
4 years ago
Can someone please help me, I’m taking a test
lawyer [7]

Answer:

Im pretty sure its none of these

8 0
3 years ago
What is the cell potential for the reaction mg(s+fe2+(aq?mg2+(aq+fe(s at 77 ?c when [fe2+]= 3.40 m and [mg2+]= 0.210 m . express
Galina-37 [17]
First, you need to calculate the standard cell potential using standard reduction potential from a textbook or online. Since Mg becomes Mg+2, magnesium is being oxidized because it is losing electrons, you need to flip its potential

Fe+2 + 2e- --> Fe                  potential= -0.44
Mg+2 + 2e- --> Mg                potential= -2.37


Cell potential= (-0.44) + (+2.37)= 1.93 V

Now, you need to use Nernst formula to get the answer. I have attached a PDF with the work.
Download pdf
8 0
3 years ago
Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water
madreJ [45]

Explanation:

The given data is as follows.

         Temperature of dry bulb of air = 25^{o}C

          Dew point = 10^{o}C = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

        Partial pressure of water vapor (P_{a}) = vapor pressure of water at a given temperature (P^{s}_{a})

Using Antoine's equation we get the following.

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}

                               = 0.17079

                   P^{s}_{a} = 1.18624 kPa

As total pressure (P_{t}) = atmospheric pressure = 760 mm Hg

                                   = 101..325 kPa

The absolute humidity of inlet air = \frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                  \frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                 = 0.00735 kg H_{2}O/ kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg H_{2}O/ kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

                 0.07 kg - 0.00735 kg

              = 0.06265 kg H_{2}O for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

                0.06265 kg \times 100

                  = 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0
3 years ago
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