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Savatey [412]
2 years ago
9

You want to clean a 500-ml flask that has been used to store a 0.9m solution. each time the flask is emptied, 1.00 ml of solutio

n adheres to the walls, and thus remains in the flask. for each rinse cycle, you pour 9.00 ml of solvent into the flask (to make 10.00 ml total), swirl to mix uniformly, and then empty it. what is the minimum number of such rinses necessary to reduce the residual concentration of 0.0001 m or below?
Chemistry
1 answer:
Doss [256]2 years ago
3 0
In this question, every time you rinse the flask, 1 ml of the solution will adhere and you add it with 9ml which will result as 10ml(1ml + 9ml). So, every time you rinse the flask it concentration will drop to 1/10 times of before. If the solution initial concentration is 0.9m, then the amount needed to rinse is:


Final concentration <= initial concentration *  (1/10) ^ rinse 
0.0001m<= 0.9m * 1/10^rinse
0.001m/ 0.9m<= 1/ 10^ rinse
0.11 * 10^-3 <=1* 10 ^ - rinse
rinse= 3

it needs at least 3x rinse to get the concentration into <<span>0.0001 m</span>
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During a lab experiment performed at STP conditions, you prepare HCl by reacting 100. ml of Cl2 gas with an excess of H2 gas.
Brrunno [24]

Answer: 19.4 mL Ba(OH)2

Explanation:

H2(g) + Cl2(g) --> 2HCl(aq) (make sure this equation is balanced first)

At STP, 1 mol gas = 22.4 L gas. Use this conversion factor to convert the 100. mL of Cl2 to moles.

0.100 L Cl2 • (1 mol / 22.4 L) = 0.00446 mol Cl2

Use the mole ratio of 2 mol HCl for every 1 mol Cl2 to find moles of HCl produced.

0.00446 mol Cl2 • (2 mol HCl / 1 mol Cl2) = 0.00892 mol HCl

HCl is a strong acid and Ba(OH)2 is a strong base so both will completely ionize to release H+ and OH- respectively. You need 0.00892 mol OH- to neutralize all of the HCl. Note that one mole of Ba(OH)2 contains 2 moles of OH-.

0.00892 mol OH- • (1 mol Ba(OH)2 / 2 mol OH-) • (1 L Ba(OH)2 / 0.230 M Ba(OH)2) = 0.0194 L = 19.4 mL Ba(OH)2

3 0
2 years ago
A sample of ammonia has a mass of 82.9 g. how many molecules are in this sample?
alina1380 [7]
The chemical formula for ammonia is NH3. So first, you need to find the molar mass of ammonia (how many grams in one mole).
N=14g
H3=3g
So one mole of NH3 is 17 grams, you can divide 82.9 grams by 17 grams to find the number of molecules. The answer should be 4.876 moles (molecules) of ammonia. Hope this helps!
8 0
2 years ago
Read 2 more answers
a metal worker uses a cutting torch that operates by reacting acetylene gas, C2H2(g), with oxygen gas, O2(g), as shown in the un
harkovskaia [24]

Answer:

650 grams

Explanation:

Given that acetylene gas reacts with oxygen to produce caobon dioxide, water and heat and the unbalnced equation is

C_2H_2(g)+O_2(g) \rightarrow CO_2(g) + H_2O(g)+ heat

Gram-formula mass of C_2H_2= 26 g/mol

So, mass of 1 mole of acelylene is 26 grams

Therefore, mass of 25 moles of acelylene=25x26=650 grams

Hence, the mass of 25 moles of acelylene is 650 grams

8 0
2 years ago
Read 2 more answers
If you start with 100 grams of hydrogen-3, how many grams will you have after 24.6 years?
svet-max [94.6K]

Answer:

The mass left after 24.6 years is 25.0563 grams

Explanation:

The given parameters are;

The mass of the hydrogen-3 = 100 grams

The half life of hydrogen-3 which is also known as = 12.32 years

The formula for calculating half-life is given as follows;

N(t) = N_0 \times \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{\frac{1}{2} }} }

Where;

N(t) = The mass left after t years

N₀ =  The initial mass of the hydrogen-3 = 100 g

t = Time duration of the decay = 24.6 years

t_{\frac{1}{2} } = Half-life = 12.32 years

N(24.6) = 100 \times \left (\dfrac{1}{2} \right )^{\dfrac{24.6}{12.32}} } = 25.0563

The mass left after 24.6 years = 25.0563 grams.

4 0
2 years ago
How much heat is absorbed/released when 25.00 g of NH3(g) reacts in the presence of excess O2(g) to produce NO(g) and H2O(l) acc
Novay_Z [31]

Answer:

429.4 kJ are absorbed in the endothermic reaction.

Explanation:

The balanced chemical equation tells us that 1168 kJ of heat are absorbed in the reaction when 4 mol of NH₃ (g) react with 5 mol O₂ (g).

So what we need is to calculates how many moles represent 25 g NH₃(g) and calculate the heat absorbed. (NH₃ is the limiting reagent)

Molar Mass NH₃  = 17.03 g/mol

mol NH₃ = 25.00 g/ 17.03 g/mol = 1.47 mol

1168 kJ /4 mol NH₃  x 1.47 mol  NH₃ =  429.4 kJ

6 0
2 years ago
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