Сильным электролитом является водный раствор: 1) co2 2) cuici2 3) c2h5oh 4) co

CuiCI2 т.к. растворимая соль

1.

дано

m(BaSO4) = 2.33 g

m(BaCL2) -?

m(H2SO4)-?

BaCL2 + H2SO4-->BaSO4+2HCL

M(BaSO4) = 233 g/mol

n(BaSO4) = m/M = 2.33 / 233 = 0.01 mol

n(BaCL2) = n(H2SO4) = n(BaSO4) = 0.01 mol

M(BaCL2) = 208 g/mol

m(BaCL2) = n*M = 0.01 * 208 = 2.08 g

M(H2SO4) = 98 g/mol

m(H2SO4) = n*M = 0.01 * 98 = 0.98 g

ответ 2.08 g , 0.98 g

2)

дано

m(Fe(OH)3) = 2.14  g

m(FeCL3) -?

m(NaOH)-?

FeCL3+ 3NaOH-->3NaCL+Fe(OH)3

M(Fe(OH)3) = 107 g/mol

n(Fe(OH)3) = m/M = 2.14 / 107 = 0.02 mol

n(FeCL3) = n(Fe(OH)3)  = 0.02 mol

M(FeCL3) = 162.5 g/mol

m(FeCL3) = n*M = 0.02 * 162.5 = 3.25 g

3n(NaOH) = n(Fe(OH)3)

n(NaOH) = 3* 0.02 = 0.06 mol

M(NaOH) = 40 g/mol

m(NaOH) = n*M = 0.06 * 40 = 2.4 g

ответ 3.25 г, 2.4 г

Объяснение:

1.

дано

m(BaSO4) = 2.33 g

m(BaCL2) -?

m(H2SO4)-?

BaCL2 + H2SO4-->BaSO4+2HCL

M(BaSO4) = 233 g/mol

n(BaSO4) = m/M = 2.33 / 233 = 0.01 mol

n(BaCL2) = n(H2SO4) = n(BaSO4) = 0.01 mol

M(BaCL2) = 208 g/mol

m(BaCL2) = n*M = 0.01 * 208 = 2.08 g

M(H2SO4) = 98 g/mol

m(H2SO4) = n*M = 0.01 * 98 = 0.98 g

ответ 2.08 g , 0.98 g

2)

дано

m(Fe(OH)3) = 2.14  g

m(FeCL3) -?

m(NaOH)-?

FeCL3+ 3NaOH-->3NaCL+Fe(OH)3

M(Fe(OH)3) = 107 g/mol

n(Fe(OH)3) = m/M = 2.14 / 107 = 0.02 mol

n(FeCL3) = n(Fe(OH)3)  = 0.02 mol

M(FeCL3) = 162.5 g/mol

m(FeCL3) = n*M = 0.02 * 162.5 = 3.25 g

3n(NaOH) = n(Fe(OH)3)

n(NaOH) = 3* 0.02 = 0.06 mol

M(NaOH) = 40 g/mol

m(NaOH) = n*M = 0.06 * 40 = 2.4 g

ответ 3.25 г, 2.4 г

Объяснение: