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4 years ago

What is simplest level at which life may exist

Chemistry

2 answers:

just olya [345]4 years ago

8 0

The simplest level at which life may exist is a cell

7nadin3 [17]4 years ago

5 0

A cell or if u wana go deeper is organells

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Which property is dependent only on the atomic structure of the compound? *

shusha [124]

I think its mass of volume

5 0

3 years ago

¿Cuántos moles de C3H8 contienen 9.25 x 10^24 moléculas?

Igoryamba

Answer:

$\large \boxed{\text{e. 15.4 mol C$_{3}$H}_{8}}$

Explanation: $\text{Moles of propane} = 9.25 \times 10^{24}\text{ molecules} \times \dfrac{\text{1 mol }}{6.022 \times\ 10^{23} \text{molecules}} = \textbf{15.4 mol}$

5 0

3 years ago

Which of the following is NOT a binary molecular compound?<br> KI <br> H2s<br> CO<br> CIF3

shutvik [7]

I believe KI is not a a binary molecule.
Your welcome

7 0

4 years ago

Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0

4 years ago

2CO + O2 --> 2CO2

GREYUIT [131]

Answer:

10 L of CO₂.

Explanation:

The balanced equation for the reaction is given below:

2CO + O₂ —> 2CO₂

From the balanced equation above,

2 L of CO reacted to produce 2 L of CO₂.

Finally, we shall determine the volume of CO₂ produced by the reaction of 10 L CO. This can be obtained as follow:

From the balanced equation above,

2 L of CO reacted to produce 2 L of CO₂.

Therefore, 10 L of CO will also react to produce 10 L of CO₂.

Thus, 10 L of CO₂ were obtained from the reaction.

3 0

3 years ago