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3 years ago

Why does a scientist form a hypothesis?

Chemistry

1 answer:

Kryger [21]3 years ago

8 0

Answer: B (to provide a statement that can be tested with an experiment

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What is the third alkali metal

andrey2020 [161]

Answer:

potassium

The third alkali metal is K (potassium). The atomic number of K (potassium) is 19. Thus, the atomic number of third alkali metal is 19

Explanation:

3 0

2 years ago

Read 2 more answers

How many moles of aspartame are present in 1.00 mg of aspartame?

Diano4ka-milaya [45]

Answer:- There are $3.40*10^-^6$ moles.

Solution:- It is a unit conversion problem where we are asked to convert mg of aspartame to moles. Aspartame is $C_1_4H_1_8N_2O_5$ and it's molar mass is 294.31 grams per mole.

mg are converted to grams and then the grams are converted to moles as:

$1.00mg Aspartame(\frac{1g}{1000mg})(\frac{1mole}{294.31g})$

= $3.40*10^-^6$ moles of aspartame

So, there would be $3.40*10^-^6$ moles of aspartame in 1.00 mg of it.

3 0

3 years ago

A balloon contains 2.0 L of air at 101.5 kPa . You squeeze the balloon to a volume of 0.25 L.

8_murik_8 [283]

<h3>Answer:</h3>

812 kPa

<h3>Explanation:</h3>

According to Boyle's law pressure and volume of a fixed mass are inversely proportional at constant absolute temperature.

Mathematically, $P\alpha \frac{1}{V}$

At varying pressure and volume;

P1V1=P2V2

In this case;

Initial volume, V1 = 2.0 L

Initial pressure, P1 = 101.5 kPa

Final volume, V1 = 0.25 L

We are required to determine the new pressure;

$P2=\frac{P1V1}{V2}$

Replacing the known variables with the values;

$P2=\frac{(101.5)(2.0L)}{0.25L}$

= 812 kPa

Thus, the pressure of air inside the balloon after squeezing is 812 kPa

8 0

2 years ago

Day and night are caused by what

Aleks04 [339]

Answer:

THE GOD ALMIGHTY!

Explanation:

6 0

3 years ago

Mixture of 120 grams of KClO3 and 300 grams of water until all of the KClO3 has dissolved. At what temperature does this occur?

lozanna [386]

240°C is required to dissolve 120 grams of KClO₃ in 300 g of water.

Explanation:

As per the solubility curve graph of different salts dissolved in 100 g of water, it can be noted that the KClO₃ gets saturated at 100 °C with the solubility of 59 g in it. But here the water is taken as 300 g and the salt is taken as 120 g. The graph shows the solubility of grams of salt in 100 g of water.

So, then we have to find what is the gram of salt present in 100 g of water for the present sample.

300 g of water contains 120 grams of salt.

Then, 1 g of water will contain $\frac{120}{300}$ g of salt.

Then, 100 g of water will contain $\frac{120}{300}*100$ of salt. This means, 100 g of water will contain 40 g of salt in it.

Then, if we check the graph, the corresponding x-axis point on the curve for the y-axis value of 40 g will give us the temperature required to dissolve the salt in water. So 80°C will be required to dissolve 40 g of salt in 100 g of water.

Then, in order to get the temperature required to dissolve 120 g in 300 g of water, multiply the temperature by 3, which will give the result as 80×3=240 °C.

So, 240°C is required to dissolve 120 grams of KClO₃ in 300 g of water.

8 0

3 years ago