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oksian1 [2.3K]
3 years ago
7

Please help do in 5 mins This system that breaks down food and absorbs food is called

Chemistry
2 answers:
Yakvenalex [24]3 years ago
8 0
A digestive system!!!
slamgirl [31]3 years ago
5 0

Answer:

the digestive system

Explanation:

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nordsb [41]
Your answer is yeast, since it is a single celled organism that can be classified into a subkingdom related to sac fungi.
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If your school said no rules wut would you do
VLD [36.1K]

Answer:

please add more info into your response

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6 0
2 years ago
A mover packs books, CDs, and DVDs into a moving box. If the box contains 6.5 kg of books, 1.5 kg of CDs, and 2.0 kg of DVDs, wh
mariarad [96]
Given: 
<span>M1 = 6.5 kg of books
</span><span>M2 = 1.5 kg of CDs
</span><span>M3 = 2.0 kg of DVDs

Required: percent by mass of each object

Solution:
First, we calculate the total mass.

M = 6.5 kg + 1.5 kg + 2.0 kg =  10 kg

Percent by mass is calculated by getting the ration of the mass of an object and the total mass multiplied by 100 to get the percent.

%M1 = 6.5 / 10 x 100 = 65%
%M2 = 1.5/10 x 100 = 15%
%M3 = 2.0/10 x 100 = 20%</span>
4 0
3 years ago
Read 2 more answers
How did the pH level and the water components level change after adding water to the battery acid?
ololo11 [35]
The pH level changed to 1.34 and the combination of the both made the mixture less acidic
the pH is 1.84, mixture had less acid and there is alot more water molecules in the mixture
the pH level is 2.13 ,, again less acidic and the water molecules has increased to 3,28 x10(25)
7 0
3 years ago
When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular
Luda [366]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is C_2H_2O_4 and C_6H_4O_2

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=1.03g

Mass of H_2O=0.14g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, \frac{12}{44}\times 1.03=0.28g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, \frac{2}{18}\times 0.14=0.016g of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = \frac{0.023}{0.00775}=2.96\approx 3

For Hydrogen  = \frac{0.016}{0.00775}=2.06\approx 2

For Oxygen  = \frac{0.00775}{0.00775}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is C_3H_{2}O_1=C_3H_2O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

n=\frac{108.10g/mol}{54g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2

Thus, the empirical and molecular formula for the given organic compound is C_3H_2O and C_6H_4O_2

6 0
3 years ago
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