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3 years ago

Which properties of matter are independent of the amount of the material?

1 answer:

3 0

Physical properties can be observed or measured without changing the composition of matter. These are properties such as mass, weight, volume, and density. Remember that density is a physical property.

You might be interested in

What is the change in density if a sample goes from 3.21 g/L to 5.43 g/mL?

Step2247 [10]

Answer:

$\Delta \rho =2.22 g/mL$

Explanation:

Hello,

In this case, since a change in science is widely known to be considered as a subtraction between the the final and initial values of two measured variables and is represented via Δ, here the final density is 5.43 g/mL and the initial one was 3.21 g/mL, therefore, the change in density is:

$\Delta \rho=\rho _f-\rho _i\\ \\\Delta \rho=5.43g/mL-3.21g/mL\\\\\Delta \rho =2.22 g/mL$

Best regards.

3 0

2 years ago

11. The back part of the cerebrum deals with hearing True False

Sunny_sXe [5.5K]

Answer:

The answer is: True

5 0

2 years ago

A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal

goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂:

$1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC$

All of the hydrogens atoms in the compound ended up becoming H₂O, so the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O:

$0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH$

Because the compound is composed only by C, H and O, the mass of O in the compound can be calculated by substraction:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C

mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H

mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

8 0

3 years ago

Help please i give brainlist

shtirl [24]

The order of the answers are as follows:
B
C
D
A

5 0

3 years ago

How much silver chloride is precipitated when 25.0 g of calcium chloride and 25.0 g of silver nitrate are mixed together in a be

Burka [1]

Answer:

21.09 g of AgCl

Explanation:

Hopefully the picture is clear and the method is understandable.

For more information go to

https://socratic.org/questions/5631d10b11ef6b4609a78ee2

7 0

2 years ago