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3 years ago

What is Alfunim second element ? Please help !!!

1 answer:

3 0

Aluminum? It is a chemical element with the symbol Al and atomic number 13. It is a silvery-white, soft, non-magnetic and ductile metal in the boron group. By mass, aluminium is the most abundant metal in the Earth's crust and the third most abundant element

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WILL GIVE BRAINIEST IF HELP Which of these is a compound

NikAS [45]

b the awnser is b its a compound of carbon C and 2 oxygon atoms O sub 2

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3 years ago

How many electrons does phosphorous (P) need to gain to have a stable outer electron level?

Brut [27]

The correct answer is 3.

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3 years ago

Read 2 more answers

How much of a 10 M solution is needed to make 1 liter of all solution?

SashulF [63]

Answer:

b is the answer 1000ml

please mark me brainlist

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3 years ago

An analytical chemist weighs out 0.055g of an unknown triprotic acid into a 250mL volumetric flask and dilutes to the mark with

Katarina [22]

Answer:

Mass of the unknown acid is 4.0g

Explanation:

The determine the molar mass of the unknown acid, the steps below can be followed

Firstly, determine the concentration of the acid, the formula below can be used;

ConcA × Va/ConcB × Vb = Na/Nb

Where ConcA is the concentration of the unknown acid

ConcB is the concentration of the NaOH base

Va is the volume of acid and Vb is the volume of base

Since, the titration was said to have reached an equivalent point, it means the number of moles of the acid (Na) was equal to the number of moles of the base (Nb) and thus both will be assumed to be 1

Thus

ConcA × 250/0.13 × 6.6 = 1/1

ConcA = 0.13 × 6.6/250

ConcA = 0.003432M

Then, determine the actual number of moles (n) of the unknown acid used,

ConcA = no of moles of acid/volume of acid (in dm³ or L)

To convert mL to L, we divide by 1000

Hence, 250ml = 0.25L

0.003432 = n/0.25

n = 0.003432 × 0.25

n = 0.01373 moles

To determine the molar mass;

n = mass/molar mass

The mass was given in the question to be 0.055g

Thus

0.01373 = 0.055/molar mass

molar mass = 0.055/0.01373

molar mass = 4.0g

5 0

3 years ago

A phosphate buffer solution (25.00 mL sample) used for a growth medium was titrated with 0.1000 M hydrochloric acid. The compone

AysviL [449]

Answer:

0,07448M of phosphate buffer

Explanation:

sodium monohydrogenphosphate (Na₂HP) and sodium dihydrogenphosphate (NaH₂P) react with HCl thus:

Na₂HP + HCl ⇄ NaH₂P + NaCl (1)

NaH₂P + HCl ⇄ H₃P + NaCl (2)

The first endpoint is due the reaction (1), When all phosphate buffer is as NaH₂P form, begins the second reaction. That means that the second endpoint is due the total concentration of phosphate that is obtained thus:

0,01862L of HCl× $\frac{0,1000mol}{L}$ = 1,862x10⁻³moles of HCl ≡ moles of phosphate buffer.

The concentration is:

$\frac{1,862x10^{-3}moles}{0,02500L}$ = 0,07448M of phosphate buffer

I hope it helps!

7 0

3 years ago