All categories

2 years ago

During what part of the experiment is the hypothesis "judged" as being rejected or supported?

Chemistry

1 answer:

Blababa [14]2 years ago

5 0

Answer:

null

Explanation:

the hypothesis is a typical statistical theory which suggest that no statistical relationship and significance exists in a set of a given single observed variable between two sets of the observed data and measured phenomena

You might be interested in

Which term best describes what Donna observed? A:dew B:fog C:mist D:rain

MariettaO [177]

I also think it’s B but not quite sure

6 0

3 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

Answer: The freezing point of solution is 2.6°C

Explanation:

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

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$

Hence, the freezing point of solution is 2.6°C

8 0

2 years ago

A system gains 687 kJ of heat, resulting in a change in internal energy of the system equal to 156 kJ. How much work is done?

Maslowich

Answer:

w = -531 kJ

1. Work was done by the system.

Explanation:

Step 1: Given data

Heat gained by the system (q): 687 kJ (By convention, when the system absorbs heat, q > 0).

Change in the internal energy of the system (ΔU°): 156 kJ

Step 2: Calculate the work done (w)

We will use the following expression.

ΔU° = q + w

w = ΔU° - q

w = 156 kJ - 687 kJ

w = -531 kJ

By convention, when w < 0, work is done by the system on the surroundings.

4 0

2 years ago

A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo

Liula [17]

Answer: The molar mass of the insulin is 6087.2 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol$

Hence, the molar mass of the insulin is 6087.2 g/mol

8 0

2 years ago

A 6.40 g sample of a compound is burned to produce 8.37 g CO_2, 2.75 g H_2O, 1.06 g N_2, and 1.23 g SO_2. What is the empirical

LuckyWell [14K]

The empirical formula :

C₁₀H₁₆N₄SO₇

<h3>Further explanation</h3>

Given

6.4 g sample

Required

The empirical formula

Solution

mass C :

= 12/44 x 8.37 g

= 2.28

mass H :

= 2/18 x 2.75 g

= 0.305

mass N = 1.06

mass S :

= 32/64 x 1.23

= 0.615

mass O = 6.4 - (2.28+0.305+1.06+0.615) = 2.14 g

Mol ratio :

= C : H : N : S : O

= 2.28/12 : 0.305/1 : 1.06/14 : 0.615/32 : 2.14/16

= 0.19 : 0.305 : 0.076 : 0.019 : 0.133 divided by 0.019

= 10 : 16 : 4 : 1 : 7

The empirical formula :

C₁₀H₁₆N₄SO₇

3 0

2 years ago