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larisa [96]
3 years ago
7

How many atoms of aluminum would be produced from the decomposition of 53.2 g of aluminum oxide?

Chemistry
1 answer:
Hoochie [10]3 years ago
3 0

Answer:

Deci trebuie sa faci unu pe trei cu Ashajfjfjgbgbxkrbcjrkjgkrbgirngkubtuxjndoskfijddjjdjf

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If a clot were made up of a mass of proteins, what change in the proteins led to the formation of a clot?
S_A_V [24]

Answer: The proteins were no longer soluble in the blood.

3 0
3 years ago
Which of the following is not an empirical formula?
PtichkaEL [24]
Option C coz it should be ( CNH4)2. Hope i cleared your doubt
7 0
3 years ago
Wich are the disadvantages of skimming oil spills?
sweet-ann [11.9K]

Oil is sucked up through wide floating heads and pumped into storage tanks. Although suction skimmers are generally very efficient, one disadvantage is that they are vulnerable to becoming clogged by debris and ice and require constant skilled observation.
4 0
3 years ago
3) During the day at 27°C a cylinder with a sliding top contains 20.0 liters
tatiyna

Answer:

T_2=12\°C

Explanation:  

Hello there!  

In this case, according to the Charles' law equation which help us to understand the directly proportional relationship between volume and temperature:

\frac{T_2}{V_2}=\frac{T_1}{V_1}  

Thus, by solving for the final temperature, T2, and making sure we use the temperatures in Kelvin, we can calculate the final temperature as shown below:

T_2=\frac{T_1V_2}{V_1}  \\\\T_2=\frac{(27+273)K*19L}{20.0L}\\\\T_2=285-273\\\\T_2=12\°C

Best regards!  

Best regards!

4 0
2 years ago
How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that
tia_tia [17]

Answer:

The correct answer is 574.59 grams.

Explanation:

Based on the given information, the number of moles of NH₃ will be,  

= 2.50 L × 0.800 mol/L

= 2 mol

The given pH of a buffer is 8.53

pH + pOH = 14.00

pOH = 14.00 - pH

pOH = 14.00 - 8.53

pOH = 5.47

The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,  

= -log (1.8 ×10⁻⁵)

= 5.00 - log 1.8

= 5.00 - 0.26

= 4.74

Based on Henderson equation:  

pOH = pKb + log ([salt]/[base])

pOH = pKb + [NH₄⁺]/[NH₃]

5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

[NH₄⁺]/[NH₃] = 10^0.73= 5.37

[NH₄⁺ = 5.37 × 2 mol = 10.74 mol

Now the mass of dry ammonium chloride required is,  

mass of NH₄Cl = 10.74 mol × 53.5 g/mol

= 574.59 grams.  

8 0
3 years ago
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