Which one of these is an element carbon carbon dioxide air water

1 answer:

8 0

Carbon is an element.
Carbon dioxide is a compound
air is a mixture
water can be both a compound/mixture depending on the type

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What do organisms have based on their functions?

mote1985 [20]

Unicellular organisms, cells perform all the basic actions and functions required for living like nutrition, respiration, reproduction etc.

In multicellular organisms, cells divide their tasks on the basis of "DIVISION OF LABOUR" to specialize in certain functions besides respiration etc. Like conduction in neurons, excitation in muscles, RBC production by bone marrow etc.

6 0

3 years ago

What mass of copper is produced when zinc is added to a solution containing 31.9g copper ii tetraoxosulphate vi

solmaris [256]

Answer:

12.7g of Cu

Explanation:

First let us generate a balanced equation for the reaction. This is illustrated below:

Zn + CuSO4 —> ZnSO4 + Cu

Molar Mass of Cu = 63.5g/mol

Molar Mass of CuSO4 = 63.5 + 32 + (16x4) = 63.5 + 32 + 64 = 159.5g/mol

From the equation,

159.5g of CuSO4 produced 63.5g of Cu.

Therefore, 31.9g of CuSO4 will produce = (31.9 x 63.5) / 159.5 = 12.7g of Cu

4 0

3 years ago

Calculate the ph of the resulting solution if 29.0 ml of 0.290 m hcl(aq) is added to

Mama L [17]

We have the given reaction as;

$HCl_{(aq)} + NaOH_{(aq)}$ ----> $NaCl_{(aq)} + H_{2}O_{(l)}$ 

Answer A) The pH will be 12.36,

We have to convert the concentrations of HCl and NaOH into moles,

So we have, n(HCl) = (0.0290 L) X (0.290 mol/L) = 8.41X $10^{-3}$ moles 

and for NaOH we have, n(NaOH) = (0.0390 L)(0.290 mol/L) = 1.13X $10^{-2}$ moles

Now, it seems NaOH is in excess, so amount remaining will be;

1.13 X $10^{-2}$ - 8.41 X $10^{-3}$ = 2.89 X $10^{-3}$ moles

Now, the total volume will become as = 0.0390 + 0.0290 = 0.068 L 

So, the concentration of [ $OH^{-}$ ] = 2.89×10ˉ³ mol / 0.068 L = 4.25 X $10^{-2}$ M

pOH = - log [ $OH^{-}$ ] = -log (4.25× $10^{-2}$ ) = 1.37

Hence, pH = 14 - pOH = 14 - 1.37 = 12.6

So the pH of the solution will be 12.6 which is basic in nature. 

Answer B) The pH will be 1.68

Now, for the given concentration we need to find moles for HCl and NaOH also;

n(HCl) = (0.0290 L)(0.290 mol/L) = 8.41 X $10^{-3}$ mol 

n(NaOH) = (0.0190 L)(0.390 mol/L) = 7.41 X $10^{-3}$ mol 

here we can see, HCl is in excess amount so the remaining will be;

8.41X $10^{-3}$ - 7.41 X $10^{-3}$ = 1.0 X $10^{-3}$ mol

Here, the total volume will be = 0.0290 + 0.0190 = 0.0480 L

So the concentration of [HCl] = 1.0 X $10^{-3}$ mol / 0.0480 L = 2.08 X $10^{-2}$ M

Which is = [H⁺] 

So, the pH = - log [ $H^{+}$ ] = -log(2.08X $10^{-2}$ ) = 1.68