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3 years ago

To get the number of atoms in a formula, add the total subscripts?

Chemistry

1 answer:

AysviL [449]3 years ago

3 0

And multiply by coefficients if there are any

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Express the following number in scientific notation: 0.000000675.

tekilochka [14]

There are 7 digits from decimal to 1st digit, and it's coming from right, so exponent will be in negative 7

In short, Your Answer would be: Option C) <span>6.75 × 10-7
</span>
Hope this helps!

6 0

4 years ago

Calculate the number of moles in 7.41 X 1017 atoms Ag.

jolli1 [7]

Answer:

1.23x10^-6 mole

Explanation:

A clear understanding of Avogadro's hypothesis proved that 1mole of any substance contains 6.02x10^23 atoms. This indicates that 1mole of Ag contains 6.02x10^23 atoms.

Now, 1f 1mole of Ag contains 6.02x10^23 atoms, then Xmol of Ag will contain 7.41x10^17 atoms i.e

Xmol of Ag = 7.41x10^17/6.02x10^23 = 1.23x10^-6 mole

4 0

3 years ago

Which of the following was not a big ideal of chemistry

ikadub [295]

Answer:

I think the answer is gravity

3 0

3 years ago

The quantitative amount of charge separation in a diatomic molecule contributes to the dipole moment of that molecule. A. True B

SOVA2 [1]

Answer:

The answer is True.

Explanation:

The sentence above is true it all adds up.

8 0

2 years ago

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2

Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

$Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}$

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

$volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL$

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

7 0

3 years ago