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Allushta [10]
3 years ago
8

A 3.45 microgram sample of Uranium has a mass of how many grams?

Chemistry
1 answer:
Ivan3 years ago
4 0
There are 3.45e-6 grams.
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Write the difference between isomerism and allotropy with one example each.
fredd [130]

Answer:

hope it helps ..

Explanation:

Allotropes can be defined as different types of compounds made out of the same single element but in different chemical formulas and different arrangements. Isomers can be defined as the chemical compounds that have a similar molecular formula but with different structural formulae. Graphite and Diamond.

7 0
3 years ago
When is a roman numeral most likely needed in the name of an ionic compound?
Sauron [17]
The answer is I’m guessing c
5 0
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The density of solid ni is 8.90 g/cm3. how many atoms are present per cubic centimeter of ni
bulgar [2K]
Density gives mass of object per volume......   Here, density is given 8.90 g/cm3   therefore, per cubic centimeter contains 8.90 g Ni.   mole of Ni = mass / atomic mass   = 8.90 / 58.6934   = 0.1516 mole     number of atoms: mole * 6.022 * 10^23   = 0.1516 * 6.022 * 10^23   = 0.9129 * 10^23   = 0.9 * 10^23 (approx.)
7 0
3 years ago
When solutions of silver nitrate and sodium chloride are mixed, a precipitation reaction occurs. What mass of precipitate can be
Bas_tet [7]

Explanation:

The given precipitation reaction will be as follows.

        AgNO_{3}(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_{3}(aq)

Here, AgCl is the precipitate which is formed.

It is known that molarity is the number of moles present in a liter of solution.

Mathematically,       Molarity = \frac{\text{no. of moles}}{\text{volume in liter}}

It is given that volume is 1.14 L and molarity is 0.269 M. Therefore, calculate number of moles as follows.

                    Molarity = \frac{\text{no. of moles}}{\text{volume in liter}}

                     0.269 M = \frac{\text{no. of moles}}{1.14 L}

                    no. of moles = 0.306 mol

As molar mass of AgCl is 143.32 g/mol. Also, relation between number of moles and mass is as follows.

               No. of moles = \frac{mass}{\text{molar mass}}

                   0.307 mol = \frac{mass}{143.32 g/mol}

                           mass = 43.99 g

Thus, we can conclude that mass of precipitate produced is 43.99 g.

                 

8 0
3 years ago
Can someone help me Balance chemical equations please
horrorfan [7]
2NaClO3 = 2NaCl + 3O2
Hopes this helps <33
4 0
2 years ago
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