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Wittaler [7]
3 years ago
12

How many grams of sugar must be dissolved in 150 mL of water to make a solution with a concentration of 0.6 g/mL?

Chemistry
1 answer:
Ksju [112]3 years ago
3 0

Answer:

250

Explanation:

all you have to do is divide them its simple division cause its by the volume by the density

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Would you sit with this person at the lunch table?
Kipish [7]

Answer:

Sure because I need more friends

7 0
3 years ago
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Please help! its due at the end of class.
Vera_Pavlovna [14]

Answer:

2.down is hypotheses and 23.across is experiment

Explanation:

akechis pancakes

5 0
3 years ago
How many moles of aluminum oxide al2o3 are in a sample with a mass of 204.0
erik [133]

Answer:

2 moles

Explanation:

Let us first start by calculating the molecular mass of Al₂O₃.

The mass of a mole of any compound is called it's molar mass. 1 molar mass 6.02 X 10²³, or Avogadro's number, of compound entities.

Say, 1 mole of Al₂O₃ has 6.02 X 10²³ of Al₂O₃ molecules/atoms. It also has 2*6.02 X 10²³ number of Al atoms and 3*6.02 X 10²³ number of O atoms.

Molecular mass of Al : 26.981539 u

Molecular mass of O: 15.999 u

Therefore, molecular mass of Al₂O₃ is:

= (2*26.981539) + (3*15.999) u

= 101.960078 u

This can be approximated to 102 u.

1mole weighs 102 u

So, 2moles will weigh 2*102 = 204 u

3 0
3 years ago
Read 2 more answers
A 50.0 g sample of liquid water at 25.0 degree C is mixed with 29.0 g of water at 45 degree C. The final temperature of the wate
kotegsom [21]

<u>Answer:</u> The final temperature of water is 32.3°C

<u>Explanation:</u>

When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid water)

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of solution 1 (liquid water) = 50.0 g

m_2 = mass of solution 2 (liquid water) = 29.0 g

T_{final} = final temperature = ?

T_1 = initial temperature of solution 1 = 25°C  = [273 + 25] = 298 K

T_2 = initial temperature of solution 2 = 45°C  = [273 + 45] = 318 K

c = specific heat of water= 4.18 J/g.K

Putting values in equation 1, we get:

50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K

Converting this into degree Celsius, we use the conversion factor:

T(K)=T(^oC)+273

305.3=T(^oC)+273\\T(^oC)=(305.3-273)=32.3^oC

Hence, the final temperature of water is 32.3°C

7 0
3 years ago
That's just the tip of the iceberg" is a popular expression you may have heard. It means that what you can see is only a small p
puteri [66]
Answer:

B 1.23 g/cc

Explanation:
For something to float on seawater, the density must be less than 1.03 g/mL. If the object sinks, the density is greater than 1.03 g/mL.

Let’s examine the answer choices. Keep in mind, the ice berg is mostly below the water level.

A. 0.88 g/cc
This is less than 1.03 g/cc, which would result in floating.

B. 1.23 g/cc
This is the best answer choice. The iceberg is mostly beneath the water, but some of it is exposed. The density is greater than 1.03 g/mL, but not so much greater that it would immediately sink.

C. 0.23 g/cc
This is less than 1.03 g/cc, which would produce floating.

D. 4.14 g/cc
This is much greater than 1.03 g/cc and the result would be sinking.
5 0
3 years ago
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