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Keith_Richards [23]
3 years ago
6

List some good ideas for Science Fair for Sixth Graders

Chemistry
1 answer:
grigory [225]3 years ago
4 0
You could test how high the soda shoots up when you drop mentos in different types of soda. i did it when i was in 6th grade
You might be interested in
Please help!!! ill mark u brainly
Musya8 [376]

Answer: I say it is A sorry if I'm incorrect

Explanation:

8 0
3 years ago
You are given two closed containers. Each containing a liquid solvent. Ammonia and table salt will dissolve in container A. Hexa
snow_tiger [21]
Ammonia and table salt dissolves in polar solvents, so A is water,
CO2 and hexane are non-polar substances, so they are going to be dissolved in non-polar solvent, so I think it is going to be carbon tetrachloride

Answer is <span>A) A - water; B - carbon tetrachloride

Table salt does not dissolve in oil and CCl4, and Br2 is too active and it is going to react with NH3.</span>
3 0
2 years ago
Read 2 more answers
Fermium-253 has a half-life of 0.334 seconds. How long will it take, in seconds, for a 96 atom sample to decay to 3 atoms?​
iris [78.8K]

Answer:3.34 seconds

Explanation:

8 0
3 years ago
What is the mass of 8.23 x 10^23 atoms of Ag
Gnom [1K]

Answer:

\boxed {\boxed {\sf Approximately \ 147 \ g\ Ag}}

Explanation:

<u>Convert Atoms to Moles</u>

The first step is to convert atoms to moles. 1 mole of every substance has the same number of particles: 6.022 ×10²³ or Avogadro's Number. The type of particle can be different, in this case it is atoms of silver. Let's create a ratio using this information.

\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}

We are trying to find the mass of 8.23 ×10²³ silver atoms, so we multiply by that number.

8.23 *10^{23} \ atoms \ Ag *\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}

Flip the ratio so the atoms of silver cancel. The ratio is equivalent, but places the other value with units "atoms Ag" in the denominator.

8.23 *10^{23} \ atoms \ Ag *\frac{1 \ mol \ Ag}{6.022*10^{23} \ atoms \ Ag}

8.23 *10^{23}  *\frac{1 \ mol \ Ag}{6.022*10^{23} }

Condense into one fraction.

\frac{8.23 *10^{23}  }{6.022*10^{23} } \ mol \ Ag

1.366655596 \ mol \ Ag

<u>Convert Moles to Grams</u>

The next step is to convert the moles to grams. This uses the molar mass, which is equivalent to the atomic mass on the Periodic Table, but the units are grams per mole.

  • Ag: 107.868 g/mol

Let's make another ratio using this information.

\frac {107.868 \ g \ Ag}{1 \ mol \ ag}

Multiply by the number of moles we calculated.

1.366655596 \ mol \ Ag*\frac {107.868 \ g \ Ag}{1 \ mol \ ag}

The moles of silver cancel out.

1.366655596 *\frac {107.868 \ g \ Ag}{1 }

1.366655596 * {107.868 \ g \ Ag}

147.4184058 \ g\ Ag

<u>Round</u>

The original measurement of atoms has 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

  • 147.<u>4</u>184058

The 4 in the tenths place tells us to leave the 7 in the ones place.

147 \ g\ Ag

8.23 ×10²³ silver atoms are equal to approximately <u>147 grams.</u>

3 0
3 years ago
Sodium carbonate (NaCO3) is sometimes used as a water-softening agent. Suppose that a worker prepares a 0.730 M solution of NaCO
sergey [27]

The molarity of a solution is the number of moles of a substance divided by the volume in liters prepared.

molarity=\frac{n}{V}, where n is number of moles and V is the volume in liters.

In order to calculate the mass of solute we need to convert the volume and molarity to moles

1.421 L solution \times\frac{0.0730 moles}{1 Lsolution}= 1.037 mol NaCO_3

Now that we have moles we use the relative formula mass of NaCO₃, We have 1 Na atom, 1 C atom and 3 O atoms, thus

M_r= (1\times 22.99) + (1\times 12.00) + (3\times 16.00)= 82.99g/mol

1.037 \times\frac{82.99g}{mol} = 86.1g

5 0
3 years ago
Read 2 more answers
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