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Oxana [17]
3 years ago
8

Which substances would be liquids at room temperature?

Chemistry
1 answer:
Vsevolod [243]3 years ago
5 0

Answer: Water,ethanol and Sulfuric acid

Explanation:

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Using the information in the passage, rank the types of galaxies by their ability to form new stars, from most to least activity
Alex Ar [27]

Answer:

The correct answer is number 2. Elliptical, Spiral, Irregular.

Explanation:

Unfortunately, you did not attach the passage. Without the passage, we do not know what its content is.

So we are going to answer this question with our own knowledge.

The types of galaxies by their ability to form new stars, from most to least activity are "Elliptical, Spiral, Irregular."

That is the way astronomers classify galaxies in scientific terms. These scientists consider that the smallest galaxies or dwarf galaxies have approximately 100 million stars. On the other hand, giant galaxies have billions or trillions of stars. That immense is our universe.

Scientists still studying the universe and the galaxies to understand the way galaxies are formed.

8 0
3 years ago
How many grams of potassium chlorate are needed to saturate 100g of water at 70 celsius? Group of answer choices 48 g 100 g 34 g
marissa [1.9K]

Answer:

34g of potassium chlorate.

Explanation:

A saturated solution is a solution that, under a temperature, has the maximum amount of solute possible. The maximum amount that a solvent can dissolve of a solute is called <em>solubility.</em>

<em> </em>

The solubility of potassium chlorate in water at 70°C is 34g/ 100g of water.

That means, to saturate 100g of water at 70°C you need yo add:

<h3>34g of potassium chlorate.</h3>
3 0
3 years ago
If 420 joules of heat energy is added to 25 grams of water at 25 degrees celcius, what will be the final temperature of the wate
natka813 [3]

Answer:

T2 = 29°C

Explanation:

Given data:

Heat added = 420 j

Mass of water = 25 g

Initial temperature = 25°C

Final temperature = ?

Solution;

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water = 4.18 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Now we will put the values.

420 j = 25 g ×4.18 j/g.°C × (Final temperature - initial temperature)

420 j = 25 g ×4.18 j/g.°C × (T2 - 25°C)

420 j = 104.5  j/°C × (T2 - 25°C)

420 j /104.5  j/°C = T2 - 25°C

4°C + 25°C = T2

T2 = 29°C

4 0
4 years ago
Do you have any ideas for a sodium poster project other than a salt shaker?
vovangra [49]

Answer:

- chips n other food items (ones with salt)

- clorox

- aleve

- baking soda

5 0
3 years ago
A radioactive isotope of potassium (k) has a half-life of 20 minutes. if a 43 gram sample of this isotope is allowed to decay fo
Ksenya-84 [330]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 43 g

m (final mass after time T) = ? (in g)

x (number of periods elapsed) = ?

P (Half-life) = 20 minutes

T (Elapsed time for sample reduction) = 80 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

80 = x*20

80 = 20\:x

20\:x = 80

x = \dfrac{80}{20}

\boxed{x = 4}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{43}{2^{4}}

m = \dfrac{43}{16}

\boxed{\boxed{m = 2.6875\:g}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

8 0
4 years ago
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