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iragen [17]
3 years ago
8

Can anyone help me with this problem :)

Chemistry
1 answer:
kiruha [24]3 years ago
4 0

Answer:

0.0003 moles

Explanation:

MgC6H8O6 = (60.0mg/1)(1 mol mg/176000mg)

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A package within package can be best defined as which of the following
Citrus2011 [14]

Answer:

double-protected

Explanation:

If a package is inside a package, the item inside the second package is double-protected, because the outer package will keep it safe.

Hope this helps!

7 0
3 years ago
Which of these is NOT a mixture?
TEA [102]
The correct answer would b C
7 0
3 years ago
Help! It’s for chemistry I attached a picture thank you!
IgorLugansk [536]

Answer:

0.15 L

Explanation:

You need to first find the volume of the container.  You can do this by dividing the mass by the density.  This will give you the mass in mL.

5.00 kg = 5,000 g

(5,000 g)/(1.00 g/mL) = 5,000 mL

5,000 mL = 5 L

Now, find the volume the seawater will take up.

(5,000 g)(1.03 g/mL) = 4854.4 mL

4854.4 mL = 4.85 L

Subtract the two volumes to find the volume that left unfilled.

5 L - 4.85 L = 0.15 L

6 0
3 years ago
A 3.25 L solution is prepared by dissolving 285 g of BaBr2 in water. Determine the molarity.
Effectus [21]

Answer:

0.295 mol/L

Explanation:

Given data:

Volume of solution = 3.25 L

Mass of BaBr₂ = 285 g

Molarity of solution = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Number of moles of solute:

Number of moles = mass/ molar mass

Molar mass of BaBr₂ = 297.1 g/mol

Number of moles = 285 g/ 297.1 g/mol

Number of moles= 0.959 mol

Molarity:

M = 0.959 mol / 3.25 L

M = 0.295 mol/L

5 0
3 years ago
The molar solubility of ba3(po4)2 is 8.89 x 10-9 m in pure water. calculate the ksp for ba3(po4)2. the molar solubility of ba3(p
Flura [38]
Answer is: The molar solubility of ba3(po4)2 is <span>6.00 x 10-39.
</span>Balanced chemical reaction: Ba₃(PO₄)₂(s) → 3Ba²⁺(aq) + 2PO₄³⁻(aq).
s(Ba₃(PO₄)₂) = 8.89·10⁻⁹ M.
[Ba²⁺] = 3s(Ba₃(PO₄)₂) = 3s.
[PO₄³⁻] = 2s.
Ksp = [Ba²⁺]³ · [PO₄³⁻]².
Ksp = (3s)³ · (2s)².
Ksp = 108s⁵.
Ksp = 108·(8.89·10⁻⁹ M)⁵.
Ksp = 108 · 5.55·10⁻⁴¹ = 6·10⁻³⁹.
3 0
3 years ago
Read 2 more answers
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