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2 years ago

How can we show the data that is collected

Chemistry

2 answers:

andreev551 [17]2 years ago

5 0

Answer:

We show the data that is collected by in this way :-

Determine What Information You Want to Collect. The first thing you need to do is choose what details you want to collect. ...

Set a Timeframe for Data Collection. ...

•Determine Your Data Collection Method. ...

Collect the Data.

Analyze the Data and Implement Your Findings

a_sh-v [17]2 years ago

4 0

Answer: Sorry I did not understand this

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What is precipitate

natima [27]

Answer:

cause something (a substance) to be deposited in solid form from a solution

Explanation:

4 0

2 years ago

What is the ph of a solution of 0.550 m k2hpo4, potassium hydrogen phosphate?

Solnce55 [7]

We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+ + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:

PH= -㏒[H+]
= -㏒(4.81x10^-7) = 6.32

3 0

3 years ago

The sea water has 8.0x10^-1 cg of element strontium. Assuming that all strontium could be recovered, how many grams of strontium

PilotLPTM [1.2K]

984 grams of strontium will be recovered from 9.84x10^8 cubic meter of seawater.

Explanation:

From the question data given is :

volume of strontium in sea water= 9.84x10^8 cubic meter

(1 cubic metre = 1000000 ml)

so 9 .84x10^8 cubic meter

$\frac{9 .84x10^8}{1000000}$ = 984 ml.

density of sea water = 1 gram/ml

from the formula mass of strontium can be calculated.

density = $\frac{mass}{volume}$

mass = density x volume

mass = 1 x 984

= 984 grams of strontium will be recovered.

98400 centigram of strontium will be recovered.

Strontium is an alkaline earth metal and is highly reactive.

4 0

3 years ago

Calculate the pH of the following simple solutions:

IceJOKER [234]

Answer :

(1) pH = 1.27

(2) pH = 13.35

(3) The given solution is not a buffer.

Explanation :

(1) 53.1 mM HCl

Concentration of HCl = $53.1mM=53.1\times 10^{-3}M$

As HCl is a strong acid. So, it dissociates completely to give hydrogen ion and chloride ion.

So, Concentration of hydrogen ion= $53.1\times 10^{-3}M$

pH : It is defined as the negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pH=-\log (53.1\times 10^{-3})$

$pH=1.27$

(2) 0.223 M KOH

Concentration of KOH = 0.223 M

As KOH is a strong base. So, it dissociates completely to give hydroxide ion and potassium ion.

So, Concentration of hydroxide ion= 0.223 M

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (0.223)$

$pOH=0.65$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-0.65=13.35$

(3) 53.1 mM HCl + 0.223 M KOH

Buffer : It is defined as a solution that maintain the pH of the solution by adding the small amount of acid or a base.

It is not a buffer because HCl is a strong acid and KOH is a strong base. Both dissociates completely.

As we know that the pH of strong acid and strong base solution is always 7.

So, the given solution is not a buffer.

5 0

2 years ago

28.7 l of propane, c3h8, are consumed in a combustion reaction. how many grams of water are produced?

lora16 [44]

The chemical reaction would be:

C3H8 + 5O2 = 3CO2 + 4H2O

For this case, we assume that gas is ideal thus in every 1 mol the volume would be 22.41 L. We calculate as follows:

28.7 L C3H8 ( 1 mol / 22.41 L ) ( 4 mol H2O / 1 mol C3H8 ) ( 18.02 g / mol ) = 92.31 g H2O produced

Hope this answers the question.

7 0

3 years ago