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Stella [2.4K]
3 years ago
12

Forming ionic compounds:

Chemistry
1 answer:
GREYUIT [131]3 years ago
4 0

Answer:

Name Formula and Charge   Name Formula and Charge

ammonium NH4+   hydroxide OH−

acetate C2H3O2−, or CH3COO− nitrate NO3−

bicarbonate (hydrogen carbonate) HCO3− nitrite NO2−

bisulfate (hydrogen sulfate) HSO4− peroxide O22−

carbonate CO32− perchlorate ClO4−

chlorate ClO3− phosphate PO43−

chromate CrO42− sulfate SO42−

cyanide CN− sulfite SO32−

dichromate Cr2O72− triiodide I3−

Explanation:

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What is the pH for a 0.10 M HCI solution at 25<br> degreesCelsius?
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Answer:

1.0

Explanation:

Hydrochloric acid is a strong acid, that is, an acid that dissociates completely, according to the following reaction.

HCl(aq) → H⁺(aq) + Cl⁻(aq)

Then, the concentration of H⁺ will be equal to the initial concentration of the acid, i.e., 0.10 M.

We can calculate the pH using the following expression.

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3 0
3 years ago
You wish to make 250. mL of 0.20 M KCl from a stock solution of 1.5 M KCl and DI water.
Nitella [24]

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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