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PilotLPTM [1.2K]
3 years ago
5

Object A: has a mass of 2g a volume of .5L

Chemistry
1 answer:
malfutka [58]3 years ago
8 0
An object will be floating in water if the density is lower and will be sink if the density is higher. Water density would be 1000g/liter or 1g/ml. To determine whether the object float or sink you need to calculate both object density.

The density of object A would be: 2g/ 0.5liter= 4g/liter
The density of object B would be: 3g/ 5liter= 0.6g/liter

It is clear that the object is much lighter than water so both objects would float.
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How many grams of k2so4 would you need to prepare 1500 g of 5.0% k2so4 solution?
son4ous [18]
Data Given:
                                  % w/w  =  5 %

                    Solution weight  =  1500 g

                       Solute weight  =  ?

Formula Used:
                            % w/w  =  (Mass of Solute / Mass of Solution) × 100

Solving for Mass of Solute,
         
                       Mass of Solute  =  (% w/w × Mass of Solution) ÷ 100

                       Mass of Solute  =  (5 × 1500 g) ÷ 100

                       Mass of Solute  =  75 g K₂SO₄
5 0
3 years ago
A chemical supply company sells a concentrated solution of aqueous h2so4 (molar mass 98 g mol−1 ) that is 50. percent h2so4 by m
trapecia [35]
Answer is: <span>the molarity of the sulfuric acid is 7.14 M.
</span>ω(H₂SO₄) = 50% ÷ 100% = 0.5.<span>
d(H</span>₂SO₄) = 1.4 g/mL.
V(H₂SO₄) = 100 mL ÷ 1000 mL/L = 0.1 L..
mr(H₂SO₄) = d(H₂SO₄) · V(H₂SO₄).
mr(H₂SO₄) = 1.4 g/mL · 100 mL.
mr(H₂SO₄) = 140 g.
m(H₂SO₄) = ω(H₂SO₄) · mr(H₂SO₄).
m(H₂SO₄) = 0.5 · 140 g.
m(H₂SO₄) = 70 g.
n(H₂SO₄) = m(H₂SO₄) ÷ M(H₂SO₄).
n(H₂SO₄) = 70 g ÷ 98 g/mol.
n(H₂SO₄) = 0.714 mol.
c(H₂SO₄) = n(H₂SO₄) ÷ V(H₂SO₄).
c(H₂SO₄) = 0.714 mol ÷ 0.1 L.
C(H₂SO₄) = 7.14 M.
8 0
2 years ago
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