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Alchen [17]
3 years ago
9

What volume of 10.0 M H2SO4 is required to prepare 4.0 L of 0.50 M H2SO4

Chemistry
2 answers:
WINSTONCH [101]3 years ago
6 0
Btw what does M stand for?
Bond [772]3 years ago
3 0
The initial concentration of H2SO4 is 10M
Given that, it has a constant number of moles.
We can say n=n
So C1xV1=C2xV2
where C is concentration and V is volume
This gives us V1=(C2V2)/C1=0.2L

So we need 0.2ml of 10M H2SO4
(We add 3.8L of distilled water to reach 4L)

M is a unit of concentration and can be written as mol/L
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How does the modern periodic table arrange elements?
katrin2010 [14]
By increasing atomic number
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3 years ago
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What r variable's :) :) :) :) :)
zhannawk [14.2K]

Answer: an element, feature, or factor that is liable to vary or change.

"there are too many variables involved to make any meaningful predictions"

Explanation:

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Which of the following salts will be less soluble in 0.10 M NaCl than it is in pure water?
MariettaO [177]

The less soluble salt : PbCl₂

<h3>Further explanation</h3>

Given

0.1 M NaCl

Required

The less soluble salt

Solution

If we see from the answer option, the salt that is more difficult to dissolve in NaCl is PbCl₂ because it has the same ion (Cl)

When PbCl₂ is dissolved in water, ionization will occur

PbCl₂ ⇒ Pb²⁺+ 2Cl⁻

So, when dissolved in NaCl, NaCl itself will be ionized

NaCl ⇒ Na⁺ + Cl⁻

Based on the principle of equilibrium, the addition of an ion (one of the ions is enlarged), the reaction will shift towards the ion that was not added. In addition to this Cl ion, the reaction will shift to the left so that the solubility of PbCl₂ will decrease (the reaction to the right decreases)

8 0
3 years ago
Gold forms face-centered cubic crystals. The atomic radius of a gold atom is 144 pm. Consider the face of a unit cell with the n
kipiarov [429]

Answer:

The length of an edge of this unit cell is 407.294 pm

Explanation:

Face centered cubic structure contains 4 atoms in each unit cell and 12 coordination number, occupying about 74% volume of the total cell. Face centered cubic structure is known for efficient use of space for atom packing.

To determine the edge length, a relationship between the radius of the atom and edge length is used.

X = R√8

Where;

X is the length of an edge of this unit cell

R is the radius of the gold atom = 144 pm = 144 X 10⁻¹² m

X = 144 X 10⁻¹²√8

X = 407.294 X 10⁻¹² m

X = 407.294 pm

Therefore, the length of an edge of this unit cell is 407.294 pm

8 0
3 years ago
A 50 L cylinder is filled with argon gas to a pressure of 10130.0 kPa at 300°C. How many moles of argon gas does the cylinder co
KengaRu [80]
In this problem, we need to use the ideal gas law. The following is the formula used in ideal gas law: PV = nRT, where n refers to the moles and R is the gas constant.

Given 
P = 10130.0 kPa 
V = 50 L
T = 300 degree celcius + 273.15 = 573.15 K
R = 8.314 L. kPa/K.mol

Solution
To get the moles which represent the "n" in the formula, we need to rearrange the equation.

PV = nRT                      PV
----    ------    --->    n = --------
 RT     RT                       RT

          10130.0 kPa  x 50 L
n= ---------------------------------------------
       8.314 L. kPa/K.mol  x 573.15 K
             506,500 
  =  ----------------------------
         4,765.17  mol K

=106.29 mol Ar

So the moles of argon gas is 106.29 moles 
8 0
3 years ago
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