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eduard
2 years ago
6

Which organisms are extremely small and use hosts as both home and food supply?

Chemistry
2 answers:
Tasya [4]2 years ago
8 0
<span><u><em>Answer:</em></u>
Parasites

<u><em>Explanation:</em></u>
Parasites are a type of <u>pathogen</u> that survive by hitchhiking a host. They usually do this as they perceive the host as type of protection for them in order to survive.

Beside this protection, the host provides these parasites with the necessary nutrition as they feed directly from the host.
<u>They can feed on different parts from the hosts including:</u>
1- the digestion as in case of tapeworm
2- brain tissue as in case of brain-eating ameoba
3- blood

Parasites' presence in the host's body can alter the functionalities and processes within the body and cause sickness.

Hope this helps :)</span>
Simora [160]2 years ago
7 0
I would say that the bacteria who live inside one's gut would constitute very small organisms that are extremely small and inhabit the gut as home and also feed off it to survive and in this case probably have a symbiotic relationship with our bodies as they help us to break down and digest food.
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space capsules operate with an oxygen content of about 34%. assuming a total pressure of 780 mm Hg in the space capsule, what is
Lemur [1.5K]

265.2 mmHg is the partial pressure of oxygen in 780 mmHg of total pressure.

Explanation:

The partial pressure of a gas is defined as the individual pressure of the gas in total mixture. In an ideal gas all the constituent gases have partial pressure some of which will give total pressure of the gas.

The partial pressure of a gas is calculated by

total pressure x mole fraction of the gas.

Mole fraction of the oxygen present is 0.34 as it is 34% of the total gas.

\frac{34}{100} = 0.34  is the mole fraction

Total pressure is given as 780 mm Hg

The partial pressure can be calculated using the above formula:

Putting the values in equation:

780 x 0.34

= 265.2 mm Hg is the partial pressure of oxygen.

7 0
2 years ago
One crystalline form of silica (SiO2) has a cubic unit cell, and from X-ray diffraction data it is known that the cell edge leng
Sav [38]

Answer:

8 Silicon atom are present in unit cell.

16 oxygen atoms are present unit cell.

Explanation:

Number of atoms in unit cell = Z =?

Density of silica = tex]2.32 g/cm^3[/tex]

Edge length of cubic unit cell = a  = 0.700 nm = 0.700\times 10^{-7} cm

1 nm=10^{-7} cm

Molar mass of Silica  = 28.09 g/mol+16.00\times 2=60.09 g/mol

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density

Z = number of atom in unit cell

M = atomic mass

(N_{A}) = Avogadro's number  

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

2.32 g/cm3=\frac{Z\times 60.09 g/mol}{6.022\times 10^{23} mol^{-1}\times (0.700\times 10^{-7}cm)^{3}}

Z = 8

1 silicon is 2 oxygen atoms. then 8 silicon atoms will be 16 oxygen atoms.

5 0
2 years ago
Two students are working together to build two models. Both models will represent the molecular structure of sodium bicarbonate,
vfiekz [6]

Answer:

Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.

<em>Note: Since no specific color was stated for oxygen atoms, the answer assigns blue colored jellybeans to represent oxygen atoms.J</em>

Explanation:

Sodium bicarbonate, NaHCO₃ is a compound composed of one atom of sodium, one atom of hydrogen, one atom of carbon and three atoms of oxygen.

Since red jellybeans represent sodium atoms, white jellybeans represent hydrogen atoms, black jellybeans represent carbon atoms and blue jellybeans represent oxygen atoms, each of the two students will require the following number of each jellybean for their model of sodium carbonate: One red jellybean, one white jellybean, one black jellybean and three blue jellybeans.

Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.

8 0
2 years ago
A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of
vlada-n [284]

Answer: The molecular of the compound is, C_2H_3O

Explanation:

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=3.095g

Mass of H_2O=1.902g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 3.095g of carbon dioxide, \frac{12}{44}\times 3.095=0.844g of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 1.902g of water, \frac{2}{18}\times 1.092=0.121g of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = (1.621)-[(0.844)+(0.121)]=0.656g

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

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles

Step 2: 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 0.041 moles.

For Carbon = \frac{0.0703}{0.041}=1.71\approx 2

For Hydrogen  = \frac{0.121}{0.041}=2.95\approx 3

For Oxygen  = \frac{0.041}{0.041}=1

Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is C_2H_3O_1=C_2H_3O

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{46.06}{43}=1

Molecular formula = (C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O

Therefore, the molecular of the compound is, C_2H_3O

6 0
2 years ago
The common constituent in all acid solutions is
Oksi-84 [34.3K]

Answer:

H+/H3O , H2O

Explanation:

The ability to be a proton donor is the Bronsted-Lowry definition of acids. The Lewis definition of an acid is an electron pair acceptor, which covers molecules liKE BF3

The ability to accept a pair of electrons is what is common to all acids, not the ability to be a proton donor.

All acid solutions contain hydronium ions (H3O+), hydroxide ions (OH-) and water molecules. Each different acid solution will then have an anion that is exclusive to that acid. For example, hydrochloric acid solution will contain all of the above and chloride ions (Cl-).

All acids contain the acidic substance dissolved in water. Water naturally dissociates to a small amount, creating hydronium and hydroxide ions. But most of the water remains as water molecules.

Then when we add an acid, like HCl, the oxygen on the water attracts the hydrogen from the HCl. The electrons in the covalent bond remain with the chlorine, giving it a negative charge and thus it becomes the chloride ion (Cl-). The hydrogen now has a positive charge and as said before, is attracted to the water (specifically the lone pair of electrons on the oxygen) to create hydronium ions.

This creates extra hydronium ions, making the solution acidic. But remember, there are still water molecules, hydroxide ions and the negative ion all in solution for all acids.

8 0
2 years ago
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