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4 years ago

A Solution has a density of 1.50g/mL How many grams are needed to obtain 10.0mL of solution

Mathematics

2 answers:

BartSMP [9]4 years ago

6 0

Answer: The required number of grams needed is 15 grams.

Step-by-step explanation: Given that a solution has a density of 1.50 g/mL.

We are to find the number of grams that are needed to obtain 10.0 mL of solution.

We will be using the UNITARY method to solve the problem.

We have

Number of grams needed o obtain 1 mL of solution = 1.50 gm.

So, number of grams needed to obtain 10.0 mL of solution is given by

$(1.50\times 10)=15~\textup{gm}.$

Thus, the required number of grams needed is 15 grams.

san4es73 [151]4 years ago

4 0

DENSITY =MASS/VOLUME

OR MASS = DENSITY X VOLUME

DENSITY =1.5 g/ml AND VOLUME = 10 ml

MASS = 1.5 x 10 = 15 g

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Resistors. An electrical engineer has two boxes of resistors, with four resistors in each box. The resistors in the first box ar

Dafna11 [192]

Answer:

A. Sample Space = S = {(9, 18),(9, 19),(9, 20),(9, 21),(10, 18),(10, 19),(10, 20),(10, 21),(11, 18),(11, 19),(11, 20),(11, 21),(12, 18),(12, 19),(12, 20),(12, 21)}

B. {(9, 18),(10, 18),(11, 18),(12, 18)}

C. {(9, 18),(9, 19),(10, 18)}

D. {(9, 18),(9, 19),(10, 18),(11, 18),(12, 18)}

E. B(complement) ∩C = {(9, 19)}

Step-by-step explanation:

The Sample Space would contain each element of the box A associated with each element of the box B.

A. Sample Space = S = {(9, 18),(9, 19),(9, 20),(9, 21),(10, 18),(10, 19),(10, 20),(10, 21),(11, 18),(11, 19),(11, 20),(11, 21),(12, 18),(12, 19),(12, 20),(12, 21)}

B. Let the Outcomes in the event B, the event that the second resistor has a resistance less than 19 be denoted by J then

J= {(9, 18),(10, 18),(11, 18),(12, 18)}

C. Let the outcomes in the event C, the event that the sum of the resistances is equal to 28 be denoted by L then

L = {(9, 18),(9, 19),(10, 18)}

D. The outcomes in B∪C contains all the elements of B and C

B∪C=J∪L= {(9, 18),(9, 19),(10, 18),(11, 18),(12, 18)}

E. B complement contains those elements of the Sample Space which are not the elements of Set B.

B complement= S-B= {(9, 19),(9, 20),(9, 21),(10, 19),(10, 20),(10, 21),(11, 19),(11, 20),(11, 21),(12, 19),(12, 20),(12, 21)}

B(complement) ∩C contains those elements of B and C which are common to both B complement and C.

B(complement) ∩C = {(9, 19)}

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3 years ago

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mylen [45]

Answer:

-65 is your answer

Step-by-step explanation:

What you do is you divide -3 to both sides.

195/-3=-65.

-65 is your answer.

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3 years ago

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3 years ago

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