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3 years ago

Which of these hypothesis is not falsifiable?

Chemistry

1 answer:

TiliK225 [7]3 years ago

8 0

Answer: it is D

Explanation:its D cause that is a opinion it is not sientificly correct

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14. How many grams of NaCl are contained in 0.350 kg of a

Lyrx [107]

Answer:

4.87g

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Mass of solution = 0.35kg

Molality = 0.238 m

Mass of NaCl =..?

Step 2:

Determination of the number of mole of NaCl in the solution.

Molality of a solution is simply defined as the mole of solute per unit kg of the solvent. It is given as:

Molality = mol of solute /mass of solvent (kg)

With the above formula, we calculate the mole of NaCl present in the solution as follow:

Molality = mol of solute /mass of solvent (kg)

0.238 = mol of NaCl /0.35

Cross multiply

mol of NaCl = 0.238 x 0.35

mol of NaCl = 0.0833 mol

Step 3:

Determination of the mass of NaCl in 0.0833 mol of NaCl.

This is illustrated below:

Number of mole NaCl = 0.0833 mol

Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol

Mass of NaCl =..?

Mass = number of mole x molar Mass

Mass of NaCl = 0.0833 x 58.5

Mass of NaCl = 4.87g

Therefore, 4.87g of NaCl is contained in the solution.

5 0

4 years ago

What is the percent by mass of carbon in c10 h14 n2?

bonufazy [111]

Add all weighs together
calculate carbons weigh
(carb weigh/total)×100

3 0

3 years ago

What is the energy change that will occur when 25.5 grams of oxygen gas (O2) react with excess methane (CH4) according to the re

AysviL [449]

Amount of CH4 is excess, so no need to worry about it
but the limiting factor is the Oxygen 

according to stranded equation, 

CH4 + 2 O2 --> CO2 + 2 H2O ΔH = -889 kJ/mol 

just by taking proportions 

(-889 kJ/mol) / 2 x 0.8 mol = - 355.6 kJ 

so i think the answer is (a)

7 0

3 years ago

Given that the molar mass of NaNO3 is 85.00 g/mol, what mass of NaNO3 is needed to make 4.50 L of a 1.50 M NaNO3 solution? Use m

Virty [35]

Ans: Mass of NaNO3 = 574 g

Given:

Volume of NaNO3, V = 4.50 L

Molarity of NaNO3, M = 1.50 M i.e. 1.50 moles/L

Molar mass of NaNO3 = 85.00 g/mol

To determine:

The mass of NaNO3

Explanation:

Molarity of a solution can be expressed as the number of moles of the solute in a liter volume of solution

Molarity of NaNO3 = moles of NaNO3/volume of solution in L

1.50 moles/L = moles/4.50 L

moles NaNO3 = 1.50 *4.50 = 6.75 moles

Now, moles of NaNO3 = mass of NaNO3/molar mass NaNO3

mass of NaNO3 = moles * molar mass =

= 6.75 moles * 85.00 g/mole

= 573.75 g = 574 g

8 0

3 years ago

Read 2 more answers

\The specific heat of aluminum is 0.21 cal g°C . How much heat is released when a 10 gram piece of aluminum foil is taken out of

DiKsa [7]

Answer: The amount of heat released is 84 calories.

Explanation:

The equation used to calculate the amount of heat released or absorbed, we use the equation:

$Q= m\times c\times \Delta T$

where,

Q = heat gained or released = ? Cal

m = mass of the substance = 10g

c = specific heat of aluminium = 0.21 Cal/g ° C

Putting values in above equation, we get:

$\Delta T={\text{Change in temperature}}=(10-50)^oC=-40^oC$

$Q=10g\times 0.21Cal/g^oC\times-40^oC$

Q = -84 Calories

Hence, the amount of heat released is 84 calories.

6 0

3 years ago