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3 years ago

9

How can constraints be used to help define the problem?

1 answer:

vitfil [10]3 years ago

6 0

Answer:

Constraints are restrictions that need to be placed upon variables

Explanation:

Constraints are restrictions (limitations, boundaries) that need to be placed upon variables used in equations that model real-world situations. It is possible that certain solutions which make an equation true mathematically, may not make any sense in the context of a real-world word problem.

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HELP. Answer as much as possible. Thnx.

blondinia [14]

Answer:4

Explanation:

6 0

3 years ago

What is the molality (m) of a solution that contains 76.5 g of KCl dissolved in 85.0 g of<br> water?

ladessa [460]

Answer:

76.5g KCl/74.55 grams per mole Kcl = x

molality= x/.085 kg H2O

Explanation:

well remember molality is moles of solute/kilograms of solvent. So it's the moles of KCl over 85 g of h20 converted into kg. if this makes sense.

5 0

3 years ago

Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.

Colt1911 [192]

Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

The amount of all elements is: (i.e formula mass of $(Ca(OH)_2)$ )

$T = 1 * Ca + 2 * H + 2 * O$

$T = 1 * 40 + 2 * 1 + 2 * 16$

$T = 74$

The amount of calcium is: (i.e formula mass of calcium)

$E = 1 * Ca$

$E = 1 * 40$

$E = 40$

So, the percentage component of calcium is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{40}{74} * 100\%$

$\%C_E = \frac{4000}{74}\%$

$\%C_E = 54.05\%$

The amount of hydrogen is:

$E = 2 * H$

$E = 2 * 1$

$E = 2$

So, the percentage component of hydrogen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{2}{74} * 100\%$

$\%C_E = \frac{200}{74}\%$

$\%C_E = 2.70\%$

Similarly, for oxygen:

The amount of oxygen is:

$E = 2 * O$

$E = 2 * 16$

$E = 32$

So, the percentage component of oxygen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{32}{74} * 100\%$

$\%C_E = \frac{3200}{74}\%$

$\%C_E = 43.24\%$

5 0

3 years ago

If i add 25 mL of water to 125 mL of a 0.15 M NaOH solution, what will the molarity of the diluted solution be ?

marshall27 [118]

1000 mL=1L

25 mL = 0.025 L
125 mL = 0.125 L

M1V1=M2V2
0.15(0.125) = M2(0.025)
0.01875 = M2(0.025)
0.75 = M2

0.75 M

6 0

3 years ago

D 6) Select True or False.

Mama L [17]

Answer:

true

Explanation:

6 0

3 years ago

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