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3 years ago

All scientists try to base their conclusions on

2 answers:

6 0

All scientists try to base their conclusions on the experiments they have conducted and they have repeated them to be sure of it.

Inessa05 [86]3 years ago

3 0

The information drawn from completing lab test and proven data from tests and hypothesis

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The following image is of a diatom. It lives in water and gets food from photosynthesis. It is encased in a hard shell made of s

Serga [27]

the correct answer is A

A diatom is a(n) Singled-celled organism because it has only one cell.

(study island)

7 0

3 years ago

A certain liquid X has a normal freezing point of 7.60 °C and a freezing point depression constant K= 6.90 °C-kg-mol. Calculate

Dmitry [639]

Answer: The freezing point of solution is -5.11°C

Explanation:

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of sodium chloride follows:

$NaCl(aq.)\rightarrow Na^{+}(aq.)+Cl^-(aq.)$

The total number of ions present in the solution are 2.

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

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

Where,

$m_{solute}$ = Given mass of solute (NaCl) = 7.57 g

$M_{solute}$ = Molar mass of solute (NaCl) = 58.44 g/mol

$W_{solvent}$ = Mass of solvent (liquid X) = 350.0 g

Putting values in above equation, we get:

$\text{Molality of }NaCl=\frac{7.57\times 1000}{58.44\times 350.0}\\\\\text{Molality of }NaCl=0.370m$

To calculate the depression in freezing point, we use the equation:

$\Delta T=iK_fm$

where,

i = Vant hoff factor = 2

$K_f$ = molal freezing point depression constant = 6.90°C/m

m = molality of solution = 0.370 m

Putting values in above equation, we get:

$\Delta T=2\times 6.90^oC/m.g\times 0.370m\\\\\Delta T=5.11^oC$

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T=\text{freezing point of water}-\text{freezing point of solution}$

$\Delta T$ = 5.11 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

$5.106^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-5.11^oC$

Hence, the freezing point of solution is -5.11°C

7 0

4 years ago

Manuel observes a chemical reaction between one atom of silicone and two atoms of oxygen which answer choice correctly list the

lawyer [7]

The answer is SiO2
Since silicon has four valence electrons and each oxygen has 2, for every 1 silicon there must be 2 oxygen to fill in both element's outer shells satisfactorily. An alternative way to figure out the chemical formula is to simply swap the charges.
Silicon is a -4 and oxygen is a -2.
Oxygen's charge is brought down into subscript and set as the number of silicons in the chemical formula, making silicon Si2. Silicon's charge is brought down to make the number of oxygens in the chemical formula O4. The formula we have currently is Si2O4. Simplify it to get the empirical formula (divide by 2) and you get SiO2.

7 0

4 years ago

Read 2 more answers

Student A performed gravimetric analysis for sulfate in her unknown using the same procedures we did . Results of her three tria

Andre45 [30]

Let us first calculate the accepted value of sulfate using the molar mass of sodium sulfate.

The molar mass of sodium sulfate is 2 Na + S + 4 O = 2 (22.99) + 32.07 + 4 (16)

Molar mass of Sodium sulfate = 142.05

Molar mass of sulfate = S + 4 (O) = 32.07 + 4 (16) = 96.07

% of sulfate in sodium sulfate = (Molar mass of SO4 / Molar mass of Na2SO4) x 100

% of sulfate = (96.07/142.05) x 100

% of sulfate = 67.6%

Theoretical value for % Sulfate is 67.6%

Percent error is calculated as

Percent Error = $\left | \frac{Experimental - Theoretical}{Theoretical} \right |$

Let us calculate mean for student A.

Mean for student A = (68.6 + 66.2 + 67.1) / 3 = 201.9/3 = 67.3 %

% error for student A = $\left | \frac{67.3 - 67.6}{67.6} \right |$

% error for student A = 0.44%

Mean for student B = (66.7 +66.6 + 66.5)/3 = 199.8 /3 = 66.6 %

% error for student B = $\left | \frac{66.6 - 67.6}{67.6} \right |$

% error for student B = 1.48%

Accuracy is the closeness of experimental values to the true value. This is defined in terms of absolute and percent errors.

Since percent error for student A is lower, we can say that student A was more accurate.

The precision of the data depends on the closeness of experimental values to each other. We can see that experimental values for student B were close to each other.

Therefore we can say that student B was more precise.

8 0

3 years ago

What are the classifications of these compounds? Are they triglycerides?

Gala2k [10]

Answer:

Yes they are triglycerides

Explanation:

Triglycerides are esters derived from glycerol and three fatty acids.

If we look at the structures closely, we will notice that the compounds are composed of glycerol and three fatty acids.

Hence the structures are triglycerides. The first structure is an unsaturated triglyceride while the second structure is a saturated triglyceride.

6 0

3 years ago