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lorasvet [3.4K]

3 years ago

9

Which best describes the act of using senses or tools to gather information? creating a hypothesis making an observation summari

zing the results recording the measurements

1 answer:

MAXImum [283]3 years ago

3 0

Answer:

making an observation

Explanation:

its the right answer

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22 °C to K (approximate room temperature)

marta [7]

Answer:

295.15 i think

Explanation:

6 0

2 years ago

One success of Johann Döbereiner's law of triads was that it organized most of the known elements into three-element groups base

jenyasd209 [6]

That statement is true

it was easy to organize 'all' the elements in a group of threes because back then they only knew about 15 - 20 elements

hope this helps

4 0

3 years ago

Read 2 more answers

What is the molality of a solution if 100.0 g of glucose (C&Hi20e) were dissolved into 750. mL of water?

hodyreva [135]

<u>Answer:</u> The molality of solution is 0.740 m.

<u>Explanation:</u>

To calculate the mass of solvent (water), we use the equation:

$Density=\frac{Mass}{Volume}$

Volume of water = 750 mL

Density of water = 1 g/mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{750mL}\\\\\text{Mass of water}=750g$

To calculate the molality of solution, we use the equation:

$Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(C_6H_{12}O_6)$ = 100.0 g

$M_{solute}$ = Molar mass of solute $(C_6H_{12}O_6)$ = 180 g/mol

$W_{solvent}$ = Mass of solvent (water) = 750 g

Putting values in above equation, we get:

$\text{Molality of }C_6H_{12}O_6=\frac{100\times 1000}{180\times 750}\\\\\text{Molality of }C_6H_{12}O_6=0.740m$

Hence, the molality of solution is 0.740 m.

6 0

3 years ago

Oxalic Acid, a compound found in plants and vegetables such as rhubarb, has a mass percent composition of 26.7% C, 2.24% H, and

blondinia [14]

Answer:

HCO₂

Explanation:

From the information given:

The mass of the elements are:

Carbon C = 26.7 g; Hydrogen H = 2.24 g Oxygen O = 71.1 g

To determine the empirical formula;

First thing is to find the numbers of moles of each atom.

For Carbon:

$=26.7 \ g\times \dfrac{1 \ mol }{12.01 \ g} \\ \\ =2.22 \ mol \ of \ Carbon$

For Hydrogen:

$=2.24 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =2.22 \ mol \ of \ Hydrogen$

For Oxygen:

$=71.1 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =4.44 \ mol \ of \ oxygen$

Now; we use the smallest no of moles to divide the respective moles from above.

For carbon:

$\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ carbon$

For Hydrogen:

$\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ hydrogen$

For Oxygen:

$\dfrac{4.44 \ mol \ of \ Oxygen}{2.22} =2 \ mol \ of \ oxygen$

Thus, the empirical formula is HCO₂

4 0

2 years ago

Which is the correct Lewis dot structure for BeH2 ?

kherson [118]

H• •Be• •H --> H:Be:H
Since there are 4 valence electrons in total, Beryllium has 2 and Hydrogen has 2. You would put the Be in the middle because there is only 1 of them.

7 0

2 years ago