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Gelneren [198K]
3 years ago
6

To make 2.00 L of 0.300 M sulfuric acid, how many mL of a 1.00 M stock solution should be used?

Chemistry
1 answer:
Tamiku [17]3 years ago
3 0
We are given with
M1 = 1.00 M

M2 = 0.300 M
V2 = 2.00 L

We are asked to get V1

Using material balance
M1 V1 = M2 V2
Substituting the given values
1.00 V1 = 0.300 M (2.00 L)
V1 = 0.600 L or 600 mL
THe volume needed is 600 mL<span />
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If 12.6 grams of iron (III) oxide reacts with 9.65 grams of carbon monoxide to produce 7.23 g of pure iron, what are the theoret
Mariana [72]
The balanced equation is Fe₂O₃ + 3 CO = 2 Fe + 3 CO₂.
Next step is to convert everything to moles.
12.6g Fe₂O₃ x (1 mol Fe₂O₃ / 159.7g Fe₂O₃) = 0.07890 mol Fe₂O₃ 
9.65g CO x (1 mol CO / 28.01g CO) = 0.3445 mol CO
The third step is to determine the limiting and excess reactants.
0.07890 mol Fe₂O₃ x (3 mol CO/1 mol Fe₂O₃) = 0.2367 mol CO
Therefore Fe₂O₃ is the limiting reagent while CO is in excess.

0.07890 mol Fe x (2 mol Fe(s) / 1 mol Fe₂O₃) = 0.1578 mol Fe(s) 
0.1578 mol Fe x (55.84g Fe / mole Fe) = 8.812g Fe is the theoretical yield
%yield = (7.23g / 8.812g) x 100% = 82.0% is the percent yield
3 0
3 years ago
1. What amount of ammonia (in moles) is produced by the reaction of 4.00 mol H2 with 3.00 mol Nz?
ArbitrLikvidat [17]
<h2>Answer:  6 moles</h2>

<h3>Explanation:</h3>

3 H₂    +    N₂   →   2 NH₃

   ↓            ↓

4 mol       3 mol

Since the moles of N₂ is the smaller of the two reactants, then N₂ is the limiting factor (the reactant that will decide how much ammonia is produced since it has the smaller amount of moles). ∴ we have to use it in calculating the number of moles of ammonia

The mole ratio of N₂ to NH₃ based on the balanced equation is 1 to 2.

∴ the moles of NH₃ = moles of N₂ × 2

                                =  3 moles × 2

                                = 6 moles

7 0
3 years ago
Three blocks are shown here pls HELPPPP:
White raven [17]

Answer: I am confident the answer is B

Explanation:

forgive me if im wrong

5 0
3 years ago
Exactly 5000 mL of air at 223K is warmed and has a new volume of 8.36 liters. What is the new temperature?
34kurt

Answer:

The new temperature is 373 K

Explanation:

Step 1: Data given

Volume air = 5000 mL = 5.0 L

Temperature = 223K

New volume = 8.36 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒V1 = the initial volume = 5.0 L

⇒T1 = the initial temperature = 223 K

⇒V2 = the new volume = 8.36 L

⇒T2 = the new temperature

5.0/223 = 8.36 /T2

T2 = 373 K

The new temperature is 373 K

7 0
3 years ago
A sample of nitrogen gas has a mass of 48.6 grams. How many N2 molecules are there in the sample? molecules Submit Answer &amp;
marysya [2.9K]

Answer:

There are 1.05  x 10²⁴ molecules in 48.6 g N₂

Explanation:

1 mol of N₂ has a mass of (14 g * 2) 28 g.

Then, 48.6 g of N₂ will be equal to (48.6 g *(1 mol/ 28 g)) 1.74 mol.

Since there are 6.022 x 10²³ molecules in 1 mol N₂, there will be

(1.74 mol *( 6.022 x 10²³ / 1 mol)) 1.05  x 10²⁴ molecules in 1.74 mol N₂ (or 48. 6 g N₂).

6 0
3 years ago
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