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enot [183]
3 years ago
11

You wish to make 250. mL of 0.20 M KCl from a stock solution of 1.5 M KCl and DI water.

Chemistry
1 answer:
Nitella [24]3 years ago
8 0

The question requires us to use the dilution formula M_iV_i=M_fV_f, where M_i and V_i are the stock concentration and volume respectively, then M_f and V_f are the dilute concentration and volume respectively.

a. C_s_t_o_c_k= 1.5 M KCI, C_d_i_l_u_t_e=0.20M KCl, V_d_i_l_u_t_e=250ml KCl

b.M_iV_i=M_fV_f\\\implies V_i= \frac{M_fV_f}{M_i} = \frac{0.20M \times 250ml}{1.5M} = 33.3\ ml

To prepare the solution 33.3ml of 1.5M KCl is diluted to a total final volume of 250ml.


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Complete question:

Write the condensed formula from left to right, starting with (CH3)x where x is a number.

See attached image for the structure formula of the compound

Answer:

(CH₃)₂CHC(CH₃)₃ named as 2,2,3-Trimethylbutane

Explanation:

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Answer:

Answer:

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