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alexandr1967 [171]
3 years ago
9

Analyze the graph to describe the solubility of barium nitrate, BaNO3. At 60°C, must be added to 1 L of water to make a saturate

d solution. At 25°C, the solubility is about . A solution that contains 600 g BaNO3 in 3 L of H2O at 70°C is . For a solution that contains 300 g BaNO3 in 3 L of H2O at 60°C to become saturated, g solute must be added.

Chemistry
1 answer:
andrew-mc [135]3 years ago
8 0

Answer:

200 g, 100 g per liter, unsaturated, 300

Explanation:

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How many moles of Na₂CO₃ required to create 9.54 liters of a 3.4 M solution
GarryVolchara [31]

Answer:

The answer to your question is 32.44 moles

Explanation:

Data

moles of Na₂CO₃ = ?

volume = 9.54 l

concentration = 3.4 M

Formula

Molarity = \frac{number of moles}{volume}

Solve for number of moles

number of moles = Molarity x volume

Substitution

Number of moles = (3.4)( 9.54)

Simplification

Number of moles = 32.44

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3 years ago
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Gwar [14]

Answer:

Explanation:

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3 years ago
How do the resonance structures for ozone, 03, differ?
Anni [7]

Explanation:

Ozone, or 03,has two major resonance structures that contribute equally to the overall hybird structure of the molecule. The two structures are equivalent from the stability standpoint, each having a positive and a negative formal charge placed on two of the oxygen atoms.

3 0
4 years ago
How many liters of 1.75 M solution could be made using 35 grams of NaCl?
dolphi86 [110]
Data:
M (molarity) = 1.75 M (mol/L)
m (mass) = 35 g
MM (molar Mass) of NaCl = 58.44 g/mol
V (volume) = ? (in liters)

Formula:
M =  \frac{m}{MM*V}

Solving:
M = \frac{m}{MM*V}
1.75 =  \frac{35}{58.44*V}
1.75*58.44V = 35
102.27V = 35
V =  \frac{35}{102.27}
\boxed{\boxed{V \approx 0.34\:L}}\end{array}}\qquad\quad\checkmark
3 0
3 years ago
In Rutherford's Gold foil experiment, were the alpha particles directed to different areas of the gold foil or only the same spo
MrRissso [65]

Answer:

See explanation

Explanation:

In the Rutherford experiment, alpha particles were directed at the same spot on a thin gold foil.

As the alpha particles hit the foil, most of the alpha particles went through the foil. In Rutherford's interpretation, most of the particles went through because the atom consisted largely of empty space.

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