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3 years ago

14

After a yellow solid is dissolved in water, the liquid appears to be uniformly yellow throughout. This yellow liquid is best cla

ssified as a(n)
a. Compound

b. Homogeneous mixture

c. Ideal Mixture

d. heterogeneous mixture

1 answer:

Vinvika [58]3 years ago

5 0

D. Heterogenous mixture

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How much work is done if an object moves 4.5 m in has 12 Newtons of force on it

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The formula of work is Work (Joules)=Force (Newtons) · distance in the direction of the force (meters), therefore its just a matter of replacing factors.

Work done = 12 · 4.5= 54 joules

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3 years ago

(a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching

stepan [7]

Answer:

a) 2.91%

b) 90.5 grams

Explanation:

Mass percentage is

mass%= $\frac{massofsoluteX100}{massofsolute+massofsolvent}$

a) the mass of NaCl = 1.50g

The mass of water = 50.0 g

Total mass =1.5+50=51.50 g

mass% = $\frac{1.5X100}{50+1.5}=2.91$ %

b) the mass percent of NaOCl = 3.62 %

It means the mass of NaOCl in 100g of bleaching solution 3.62 g

so mass of NaOCl in 2.50kg = 0.0905 kg = 90.5g

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3 years ago

What does the word renal vein mean

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a short thick vein that is formed in each kidney by the convergence of the interlobar veins, leaves the kidney through the hilum, and empties into the inferior vena cava.

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4 years ago

A 20g sample of iron at a temperature of 120oC is placed into a container of water. There are 300 milliliters of water in the co

galben [10]

Answer:

30.63 °C will be the final temperature of the water.

Explanation:

Heat lost by iron will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of iron = $m_1=20 g$

Specific heat capacity of iron = $c_1=0.444 J/g^oC$

Initial temperature of the iron = $T_1=120^oC$

Final temperature = $T_2$ =T

$Q_1=m_1c_1\times (T-T_1)$

Volume of water = 300 ml

Density of water = 1 g/mL

Mass of water= $m_2=300 mL\times 1 g/mL = 300 g$

Specific heat capacity of water= $c_2=4.184 J/g^oC$

Initial temperature of the water = $T_3=30^oC$

Final temperature of water = $T_2$ =T

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

we get, T = 30.63°C

30.63 °C will be the final temperature of the water.

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Answer:

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