25 mL of 3.0 M HNO_3 are mixed with 75 mL | vedclass

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(C) The molarity of the final mixture is calculated using the formula: $M_1V_1 + M_2V_2 = M_{final}V_{final}$.Given: $M_1 = 3.0 \ M$,$V_1 = 25 \ mL$,$

Vedclass · Accepted Answer

$3.75$

Vedclass · Answer

(C) The molarity of the final mixture is calculated using the formula: $M_1V_1 + M_2V_2 = M_{final}V_{final}$.Given: $M_1 = 3.0 \ M$,$V_1 = 25 \ mL$,$M_2 = 4.0 \ M$,$V_2 = 75 \ mL$.Total volume $V_{final} = V_1 + V_2 = 25 \ mL + 75 \ mL = 100 \ mL$.Substituting the values: $(3.0 	imes 25) + (4.0 	imes 75) = M_{final} 	imes 100$.$75 + 300 = M_{final} 	imes 100$.$375 = M_{final} 	imes 100$.$M_{final} = \frac{375}{100} = 3.75 \ M$.

$25 \ mL$ of $3.0 \ M \ HNO_3$ are mixed with $75 \ mL$ of $4.0 \ M \ HNO_3$. If the volumes are additive,the molarity of the final mixture would be ......... $M$.

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