All categories

3 years ago

What would happen to the block if it had a density of 0.500 kg/L and was placed in the same 100.0 L tank of water?

Chemistry

2 answers:

Bezzdna [24]3 years ago

8 0

Answer : The correct option is, It would float.

Explanation :

As we know that the density of water is, 1 kg/L.

The given density of block is, 0.500 kg/L.

When the density of a substance will be greater than the density of water then the substance will sink in the water.
When the density of a substance will be lesser than the density of water then the substance will float in the water.

As per question we conclude that, the density of substance (block) is less than the density of water. So, the block will float in the tank of water.

Hence, the correct option is, It would float.

raketka [301]3 years ago

6 0

I’m not sure if there was important information in the question before this one, but the answer based on the info I have is B.

The density of water is 1kg/L. Since the density of the block is less, it will float.

You might be interested in

You are working the grill at a restaurant and are having trouble getting the grill lit through electrical means. You light a mat

never [62]

Answer:

in this situation I would a little bold

Explanation:

first I don't know what extinguisher I would use pretty much any that helps with fires. I'll back people up, take the hood and put it on the small fire that way it will light out more and if I open the hood and there still a little fire I would use the extinguisher and no one gets hurt :)

6 0

3 years ago

HELP ASAP, TIMED ASSINGMENT

Oksanka [162]

Answer:

All of the physical and biological factors in it's environment, I think this is the answer, but I might be wrong.

Explanation:

I learned it in my school's Science class

8 0

3 years ago

Read 2 more answers

An unknown radioactive substance has a half-life of 3.20hours . If 46.2g of the substance is currently present, what mass A0 was

iogann1982 [59]

Answer:

a) a0 was 46.2 grams

b) It will take 259 years

c) The fossil is 1845 years old

Explanation:

An unknown radioactive substance has a half-life of 3.20hours . If 46.2g of the substance is currently present, what mass A0 was present 8.00 hours ago?

A = A0 * (1/2)^(t/h)

⇒ with A = the final amount = 46.2 grams

⇒ A0 = the original amount

⇒ t = time = 8 hours

⇒ h = half-life time = 3.2 hours

46.2 = Ao*(1/2)^(8/3.2)

Ao = 261.35 grams

Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 34.0% of an Am-241 sample to decay?

t = (ln(0.66))-0.693) * 432 = 259 years

It will take 259 years

A fossil was analyzed and determined to have a carbon-14 level that is 80% that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?

t = (ln(0.80))-0.693) * 5730 = 1845

The fossil is 1845 years old

7 0

3 years ago

A student heats a sample of hydrate once, and the mass of the sample and the evaporating dish is 16.428 g. After a second heatin

jolli1 [7]

Answer:

12.371 g

Explanation:

Given :

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g$

$m_{evaporating\ dish}+m_{First\ heated\ sample}=16.428\ g$

$m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g$

Mass of salt hydrate:

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g$

$m_{Hydrate\ sample}=25.637-m_{evaporating\ dish}\ g=25.637-1.135\ g=24.502\ g$

Mass of salt anhydrous:

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g$

$m_{Second\ heated\ sample}=m_{salt\ anhydrous}=13.266-m_{evaporating\ dish}\ g=13.266-1.135\ g=12.131\ g$

Mass of water:

$m_{water}=m_{Hydrate\ sample}-m_{salt\ anhydrous}=24.502-12.131\ g=12.371\ g$

$m_{water}=12.371\ g$

4 0

3 years ago

Using the balanced equation for the next few questions: 4 Fe(s) + 30 (9) - 2Fe2O3(s)

stira [4]

Answer:

0.075 moles of iron oxide would be produced by complete reaction of 0.15 moles of iron.

Explanation:

The balanced reaction is:

4 Fe + 3 O₂ → 2 Fe₂O₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Fe: 4 moles
O₂: 3 moles
Fe₂O₃: 2 moles

You can apply the following rule of three: if by stoichiometry 4 moles of Fe produce 2 moles of Fe₂O₃, 0.15 moles of Fe produce how many moles of Fe₂O₃?

$moles of Fe_{2} O_{3} =\frac{0.15 moles of Fe*2 moles of Fe_{2} O_{3} }{4 moles of Fe}$

moles of Fe₂O₃= 0.075

0.075 moles of iron oxide would be produced by complete reaction of 0.15 moles of iron.

5 0

2 years ago