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Arturiano [62]
3 years ago
5

How do scientists write very large numbers?

Chemistry
1 answer:
sineoko [7]3 years ago
4 0

Your best bet is B or D. Most likely B

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Which of the four solutions is the best buffer against addition of acid or base?
Vera_Pavlovna [14]

b. Na2HPO4 + NaH2PO4.

A <em>buffer </em>is a solution of a weak acid and its conjugate base. The weak acid is H2PO4^(-) and its conjugate base is HPO4^(2-).

All the other options are incorrect because they consist of only a single component.

8 0
3 years ago
What is the temperature of the mixture formed of 100 g of water with temperature 15c and 250g of water with temperature 50c in a
frez [133]

Hello I am a American person

8 0
4 years ago
What subatomic particle was discovered with the use of a cathode-ray tube
slamgirl [31]

Answer:

The correct answer is: Electron

Explanation:

Electron (e⁻) is the negatively charged subatomic particle that has an electric charge of -1e (−1.6022 ×10⁻¹⁹ C). It was discovered by the English physicist, Sir Joseph John Thomson (J. J. Thomson) in the year 1897 by the <em>cathode ray experiment</em>.

In the cathode ray experiment, <u>Sir Joseph John Thomson used the </u><em><u>cathode ray tubes</u></em><u> </u><u>to conclude the presence of negatively charged small subatomic particle, known as electrons, in all atoms.</u>

3 0
3 years ago
A binding protein binds to a ligand l with a kd of 400 nm. what is the concentration of ligand when y is (a) 0.25, (b) 0.6, (c)
lilavasa [31]

Hey there!:

The fractional saturation y is defined as :

y =  [ L ] / Kd + [ L ]

where :

[ L ] = concentration of binding ligand

Kd = 400 nm

3 0
4 years ago
Assume that 50.0mL 50.0mL of 1.0MNaCl(aq) 1.0MNaCl(aq) and 50.0mL 50.0mL of 1.0M AgNO 3 (aq) 1.0MAgNO3(aq) were combined. Accord
S_A_V [24]

Answer:

The amount of precipitate formed would 7.175 grams of silver chloride.

Explanation:

Moles (n)=Molarity(M)\times Volume (L)

Moles of NaCl = n

Volume of NaCl solution = 50.0 mL = 0.050 L

Molarity of the hydrogen peroxide = 2.0 M

n=2.0 M\times 0.050 L=0.100 mol

Moles of silver nitarte = n'

Volume of silver  nitrate solution = 50.0 mL = 0.050 L

Molarity of the silver nitrate = 1.0 M

n'=1.0 M\times 0.050 L=0.050 mol

NaCl(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NaNO_3(aq)

According to reaction, 1 mole of of silver nitrate reacts with 1 mole of NaCl. Then 0.050 mole of silve nitrate will :

\frac{1}{1}\times 0.050 mol=0.050 mol of NaCl

This means that silver nitrate is in limiting amount and NaCl is in excessive amount.

So, the amount of AgCl depends upon amount of silver nitrate.

According to reaction, 1 mole of silver nitrate gives 1 mole of AgCl.

Then 0.050 moles of silver nitrate will give;

\frac{1}{1}\times 0.050 mol=0.050 mol of AgCl

Mass of 0.050 moles of AgCl ;

0.050 mol\times 143.5 g/mol=7.175 g

The amount of precipitate formed would 7.175 grams of silver chloride.

8 0
3 years ago
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