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A 12.0% sucrose solution by mass has a density of 1.05 gem, what mass of sucrose is present in a 32.0-mL sample of this solution

natulia [17]

Answer:

Option C. 4.03 g

Explanation:

Firstly we analyse data.

12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.

Density is the data that indicates grams of solution in volume of solution.

We need to determine, the volume of solution for the concentration

Density = mass / volume

1.05 g/mL = 100 g / volume

Volume = 100 g / 1.05 g/mL → 95.24 mL

Therefore our 12 g of solute are contained in 95.24 mL

Let's finish this by a rule of three.

95.24 mL contain 12 g of sucrose

Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g

7 0

3 years ago

How does a substances temperature change when the average kinetic energy of its particles increases ? When the average kinetic e

Novay_Z [31]

Answer:

The temperature of a substance when the average kinetic energy of its particles increases and decreases when the average kinetic energy decreases.

Explanation:

Atoms and molecules are in constant motion. Kinetic energy is a form of energy, known as energy of motion. Kinetic energy is a form of energy, known as energy of motion. The kinetic energy of an object is that which is produced due to its movements, which depends on its mass (m) and speed (v).

Temperature refers to a quantity used to measure the kinetic energy of a system. That is, temperature is defined as an indicator of the average kinetic energy of the particles in a body.

So, since temperature is a measure of the speed with which they move, the higher the temperature the faster they move.

Finally, the temperature of a substance when the average kinetic energy of its particles increases and decreases when the average kinetic energy decreases.

8 0

3 years ago

What coefficient should be placed before 02? 2S+ __02 → 2S03

Trava [24]

Explanation:

$\tt{2S+6O_2\implies{2SO_3} }$

5 0

2 years ago

A chemistry student weighs out of sulfurous acid , a diprotic acid, into a volumetric flask and dilutes to the mark with distill

nadezda [96]

The question is incomplete, here is the complete question:

A chemistry student weighs out 0.104 g of sulfurous acid, a diprotic acid, into a 250.0 mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

Answer: The volume of NaOH needed is 36.2 mL

Explanation:

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

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

Given mass of sulfurous acid = 0.104 g

Molar mass of sulfurous acid = 82 g/mol

Volume of solution = 250 mL

Putting values in above equation, we get:

$\text{Molarity of sulfurous acid}=\frac{0.104\times 1000}{82\times 250}\\\\\text{Molarity of sulfurous acid}=5.07\times 10^{-3}M$

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=5.07\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.0700M\\V_2=?mL$

Putting values in above equation, we get:

$2\times 5.07\times 10^{-3}\times 250.0=1\times 0.0700\times V_2\\\\V_2=\frac{2\times 5.07\times 10^{-3}\times 250}{1\times 0.0700}=36.2mL$

Hence, the volume of NaOH needed is 36.2 mL

5 0

3 years ago

Mercury chloride is a commercial fungicide. If the molar mass is 470 g/mol and the percent composition is 85.0% Hg and 15.0% Cl,

ElenaW [278]

Data: molar mass 470 g/mol

Percent composition:

Hg = 85.0%
Cl = 15.0%

Solution:

1) Convert % to molar ratios

A. Base: 100 g

=> Hg = 85.0 g / 200.59 g/mol = 0.4235 mol

Cl = 15.0 g / 35.45 g/mol = 0.4231 mol

B. divide by the higher number and round to whole number

Hg = 0.4325 / 0.4231 = 1.00

Cl = 0.4231 / 0.4231 = 1.00

=> Empirical formula = Hg Cl

2) Find the mass of the empirical formula:

HgCl: 200.59 g/mol + 35.45 g/mol = 236.04

3) Determine how many times is the empirical mass contained in the molecular mass:

470 g/mol / 236.04 = 1.99 ≈ 2

=> Molecular formula = Hg2 Cl2.

Answers:

Empirical formula HgCl
Molecular Formula Hg2Cl2

6 0

3 years ago