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nalin [4]
3 years ago
11

If you use 1 mole of NaOH, how much NaAl(OH)4 is produced

Chemistry
1 answer:
abruzzese [7]3 years ago
5 0
Answer:
             81.97 g of NaAl(OH)₄

Solution:
              The reaction for the preparation of Sodium Aluminate from Aluminium Metal and NaOH is as follow,

                 2 NaOH  +  2 Al  +  6 H₂O    →    2 NaAl(OH)₄  +  3 H₂

According to this equation,

          2 Moles of NaOH produces  =  163.94 g (2 mole) of NaAl(OH)₄
So,
       1 Mole of NaOH will produce  =  X g of NaAl(OH)₄

Solving for X,
                     X  =  (1 mol × 163.94 g) ÷ 2 mol

                     X  =  81.97 g of NaAl(OH)₄
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Helpppppp i hate chemistry
BigorU [14]

Answer:

The first option: Strontium Fluorate.

Explanation:

because Fluorine and oxygen combines to make fluorate, Strontium stays the same.

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8 0
3 years ago
How many moles of H2O2 are needed to react with 1.07 moles of N2H4?
I am Lyosha [343]

Answer:

2.14 moles of H₂O₂ are required

Explanation:

Given data:

Number of moles of H₂O₂ required = ?

Number of moles of N₂H₄ available = 1.07 mol

Solution:

Chemical equation:

N₂H₄  +   2H₂O₂       →   N₂ +  4H₂O

now we will compare the moles of H₂O₂ and N₂H₄

                          N₂H₄     :      H₂O₂  

                            1           :        2

                            1.07      :         2×1.07 = 2.14 mol

                   

6 0
3 years ago
Which units are used to measure both velocity and speed? Select three options
Ede4ka [16]

Answer and Explanation:

The options aren't listed in your question, but here are some units that are regularly and normally used (in the classroom and in the outside world):

(The SI unit of distance and displacement is the meter. The SI unit of time is the second.)

<u>Meters per Second (m/s)</u>

kilometers per hour (km/hr)

kilometers per second (km/sec)

To find the average speed, you do distance divided by time.

To find the average velocity, you do the final position minus the initial position, divided by the final time minus the initial time.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

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8 0
3 years ago
There are three factors in which natural selection occurs. Select one statement from the list below that is a true statement abo
azamat

Answer: There is no list in your question. But the three factors in which natural selection occurs are:

1. A struggle for existence,

2. Variation and

3. Inheritance

Explanation:

Darwin’s theory of natural selection is the way in which organisms change overtime.

These changes allow organisms to become better adapted to their environment and increase their chances for survival.

The three major factors through which natural selection occurs are:

1. A struggle for existence,

2. Variation and

3. Inheritance

Darwin’s theory of natural selection will occur if the three conditions above are met and unless these factors are in place, natural selection will not occur.

8 0
3 years ago
The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = \frac{0.570 g}{122.12 g/mol}=0.004667 mol

Energy released by 0.004667 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

Q=C\times \Delta T

15,066.8 J=C\times 2.053^oC

C=7,338.92 J/^oC

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = \frac{2.900 g}{180.16 g/mol}=0.016097 mol

Energy released by the 0.016097 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

44,749.1 J=7,338.92 J/^oC\times \Delta T'

\Delta T'=6.097^oC

The temperature change from the combustion of the glucose is 6.097°C.

6 0
3 years ago
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