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Novosadov [1.4K]
3 years ago
8

How many grams of k2so4 would you need to prepare 1500 g of 5.0% k2so4 solution?

Chemistry
1 answer:
son4ous [18]3 years ago
5 0
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
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4 0
3 years ago
What is the total mass in grams of 0. 75 mole of so2.
Bas_tet [7]

Answer:

48.075g(or 48g in correct sig figs)***

Explanation:

\frac{0.75mol}{1} *\frac{64.1g}{1 mol}=48.075g

*64.1g is the mass of SO2 which is calculated by simply taking the mass of sulfur and oxygen(but doubling it since there are two oxygens) and adding them together(32.1+2x16.0)

**btw the mol units cancel because of dimensional analysis in case anyone was wondering why

***if your teacher is like mine and specifically wants your answer in correct sig figs, use the answer in parentheses as the original problem only has 2 sig figs

7 0
2 years ago
Which statement is true of a catalyst?
chubhunter [2.5K]

<u><em>Answer:</em></u>

Correct option is D.

It accelerates the reaction rates of a mixture.

<u><em>Explanation:</em></u>

It is used to speed up a reaction by lowering the activation energy.Catalysis is the backbone of many industrial processes, which use chemical reactions to turn raw materials into useful products.

<u><em>Types</em></u>

There are two types of catalyst (1) Homogeneous (2) Heterogeneous

In a heterogeneous reaction, the catalyst is in a different phase from the reactants. In a homogeneous reaction, the catalyst is in the same phase as the reactants.

6 0
3 years ago
The number of atoms present in 2.5 mole of triatomic gas is equivalent to ​
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Answer: 1.41725 X 10^{24} atoms

Explanation:

One male of any gas containing no. of atoms$=6.023\times10^{23}$

2.5 moles of gas containing no. of atoms$=2.5\times6.023\times10^{23}$$=15.0575\times10^{23}$

The given gas is triatomic gas. Hence

No. of atoms$=3\times15.075\times10^{23}$

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