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3. HeBsiskil B mpupautemistx koopinmar COCTABH.III: f_(x)=0,22M;f_(y)=-0,16M A6co.110THast HeBsiska paBila a) 0,24 M: b) 0,15 M: c) 0,62 M: d) 0,27 M.

Вопрос

3. HeBsiskil B mpupautemistx koopinmar COCTABH.III: f_(x)=0,22M;f_(y)=-0,16M A6co.110THast
HeBsiska paBila
a) 0,24 M:
b) 0,15 M:
c) 0,62 M:
d) 0,27 M.

3. HeBsiskil B mpupautemistx koopinmar COCTABH.III: f_(x)=0,22M;f_(y)=-0,16M A6co.110THast HeBsiska paBila a) 0,24 M: b) 0,15 M: c) 0,62 M: d) 0,27 M.

Решения

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мастер · Репетитор 5 лет

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To find the concentration of the solution, we can use the formula:<br /><br />\[ C = \frac{m}{V} \]<br /><br />where:<br />- \( C \) is the concentration of the solution,<br />- \( m \) is the mass of the solute, and<br />- \( V \) is the volume of the solution.<br /><br />Given that the concentration of the solution is 0.22M and the volume is 0.16M, we can substitute these values into the formula:<br /><br />\[ C = \frac{0.22M}{0.16M} \]<br /><br />Simplifying the equation, we get:<br /><br />\[ C = 1.375M \]<br /><br />Therefore, the concentration of the solution is 1.375M.
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