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sesenic [268]
3 years ago
6

Will give Brainliest!!! Will award answer! The following reaction shows the products when sulfuric acid and aluminum hydroxide r

eact.
2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

The table shows the calculated amounts of reactants and products when the reaction was conducted in a laboratory.

Initial Mass and Yield
Sulfuric Acid Aluminum Hydroxide
Initial Amount of Reactant 40 g 45 g
Theoretical Yield of Water from Reactant 14.69 g 31.15 g
Thank you for helping me learn about chemistry and apply skills.

What is the approximate amount of the leftover reactant? (4 points)
20.89 g of sulfuric acid
22.44 g of sulfuric acid
21.22 g of aluminum hydroxide
23.78 g of aluminum hydroxide
Chemistry
2 answers:
kap26 [50]3 years ago
8 0

Answer;

23.78g of aluminum hydroxide

Explanation:

DochEvi [55]3 years ago
6 0

Answer:

23.78 Grams of Aluminum Hydroxide

Explanation:

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AlekseyPX
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Balance the equation<br> N2O5 + H2 -&gt; NH3 + H2O
Lubov Fominskaja [6]

Answer:

I hope this is it. I'm not really sure.

4 0
3 years ago
In the early 1960s, radioactive strontium-90 was released during atmospheric testing of nuclear weapons and got into the bones o
Advocard [28]

Answer:

52.54 %

Explanation:

Half life = 29 years

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac {ln\ 2}{29.0}\ {years}^{-1}

The rate constant, k = 0.023902 hour⁻¹

From 1964 to 1991:

Time = 27 years

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

So,  

\frac {[A_t]}{[A_0]}=e^{-0.023902\times 27}

\frac {[A_t]}{[A_0]}=0.5245

<u>The strontium-90 remains in the bone = 52.54 %</u>

5 0
3 years ago
A gaseous mixture consisting of nitrogen, argon, and oxygen is in a 3.5-L vessel at 25C. Determine the number of moles of oxygen
Step2247 [10]

Answer:

Number of moles of oxygen = 0.037  mol

Explanation:

Given data:

Total pressure = 98.5 KPa

Partial pressure of nitrogen = 22.0 KPa

Partial pressure of argon = 50.0 KPa

Volume = 3.5 L

Temperature = 25°C (25+273= 298K)

Number of moles of oxygen = ?

Solution:

Total pressure = P(N₂) + P(O₂) + P(Ar)

98.5 KPa = 22.0 KPa +P(O₂) + 50.0 KPa

98.5 KPa = 72.0 KPa +P(O₂)

P(O₂)  = 98.5 KPa - 72.0 KPa

P(O₂)  = 26.5 KPa

KPa to atm:

26.5 KPa/ 101 = 0.262 atm

Number of moles of oxygen:

PV = nRT

n = PV/RT

n = 0.262 atm × 3.5 L / 0.0821 atm.L/mol.K  × 298 K

n = 0.917atm.L /24.47atm.L/ mol

n = 0.037  mol

6 0
3 years ago
Using the first volume and pressure reading on the table as V1 and P1, solve for the unknown values in the table below. Remember
maw [93]
Hope this helps. I provided step by step in the picture below if you want to see how I got these answers.
A= 1.0L
B= 0.50atm
C= 0.60atm
D= 4.0L

8 0
3 years ago
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