All categories

3 years ago

How do scientist classify newly discovered organisms into domains and kingdoms? (6th Grade)

Chemistry

1 answer:

olga55 [171]3 years ago

5 0

Answer: Scientists look for cell types, the ability to make food, number of cells in their bodies.

Explanation: Scientists look for cell types, the ability to make food, number of cells in their bodies when placing newly-discovered organisms.

You might be interested in

Identify two ways that the rate of a chemical reaction could be increased

Svetlanka [38]

Increase the Surface Area of the Reactants. Increasing the surface area of the reactants increases the rate of the reaction. More surface area means more collisions of the reactant molecules and an increased rate of the reaction. This occurs when reactants are made to react in powdered form

7 0

3 years ago

Which of the following statements about tolerance range is true?

vovikov84 [41]

A. Tolerance range is different for different organisms.

6 0

3 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

3 years ago

Need help !!!!! ASAP

gavmur [86]

<h2>Hello!</h2>

The answer is:

The new temperature will be equal to 4 K.

$T_{2}=4K$

We are given the volume, the first temperature and the new volume after the gas is compressed. To calculate the new temperature after the gas was compressed, we need to use Charles's Law.

Charles's Law establishes a relationship between the volume and the temperature at a gas while its pressure is constant.

Now, to calculate the new temperature we need to assume that the pressure is kept constant, otherwise, the problem would not have a solution.

From Charle's Law, we have:

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

So, we are given the following information:

$V_{1}=500mL\\T_{1}=20K\\V_{2}=100mL$

Then, isolating the new temperature and substituting the given information, we have:

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

$T_{2}=\frac{T_{1}}{V_{1}}*V_{2} \\$

$T_{2}=\frac{20.00K}{500mL}*100mL\\$

$T_{2}=4K$

Hence, the new temperature will be equal to 4 K.

$T_{2}=4K$

Have a nice day!

7 0

3 years ago

What are some ways that organelles are similar to organ system?

Vaselesa [24]

Explanation:

Like organs in an organism, each organelle has a specific function in a cell. All of the organelles work together to carry out the functions of the cell as a whole, just as organs do to an organism. For example, mitochondria are the sites of cellular respiration; they provide energy the cell

4 0

3 years ago