What happens to the kinetic energy of a ball if its velocity is increased to twice its original value?

vazorg [7]

Hey wademcelroy2005, thanks for submitting your question to Brainly!

The answer to your question is B:Radiation moves from a warmer object to a cooler object

4 0

3 years ago

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if you are told to get 100 mL of stock solution to use to prepare smaller size sample for an experiment, which piece of glasswar

fgiga [73]

Answer:

A beaker

Step-by-step explanation:

Specifically, I would use a 250 mL graduated beaker.

A beaker is appropriate to measure 100 mL of stock solution, because it's easy to pour into itscwide mouth from a large stock bottle.

You don't need precisely 100 mL solution.

If the beaker is graduated, you can easily measure 100 mL of the stock solution.

Even if it isn't graduated, 100 mL is just under half the volume of the beaker, and that should be good enough for your purposes (you will be using more precise measuring tools during the experiment).

6 0

2 years ago

What is the mass if 1.72 moles of sodium nitrate

alexgriva [62]

Answer:

146 g

Explanation:

Step 1. Calculate the molar mass of NaNO₃

Na = 22.99

N = 14.01

3O = 3 × 16.00 = 48.00

Total = 85.00 g/mol

Step 2. Calculate the mass of NaNO₃

Mass of NaNO₃ = 1.72 × 85.00/1

Mass of NaNO₃ = 146 g

8 0

3 years ago

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What are the two different ions present in the compound al(no3)3?

Aleks [24]

$Al^{3+}$ and $NO_3^{-}$

Aluminum nitrate is an ionic compound, meaning that it is made up cations and anions. By convention when writing the formula for an ionic compounds the cation part shall be placed in front the anions. Thus the cation would contain aluminum and the anion shall be a polyatomic nitrate ion consisting of nitrogen and oxygen. The bracket with subscript 3 attached resembles the ratio between the two species, aluminum ions and nitrate ions. For each aluminum ion in the compound there shall be three corresponding nitrate ions.

Aluminum form ions of charge +3 whereas nitrate ions are of charge -1. Write the charge of the ion as the superscript in reverse (i.e., write $3+$ as in $Al^{3+}$ and $-$ in $NO_3^{-}$ , omitting any 1.)

5 0

3 years ago

How many atoms are in 25.00 g of B.

iren2701 [21]

Answer:

$\boxed {\boxed {\sf 1.393 *10^{24} \ atoms \ B}}$

Explanation:

1. Convert Grams to Moles

Use the molar mass (found on the Periodic Table) to convert from grams to moles.

Boron (B): 10.81 g/mol

Use this value as a ratio.

$\frac {10.81 \ g \ B }{1 \ mol \ B}$

Multiply by the given number of grams.

$25.00 \ g \ B *\frac {10.81 \ g \ B }{1 \ mol \ B}$

Flip the ratio so the grams of boron cancel out.

$25.00 \ g \ B *\frac {1 \ mol \ B }{10.81 \ g \ B}$

$25.00 *\frac {1 \ mol \ B }{10.81 }$

$\frac {25.00 \ mol \ B }{10.81 }=2.312673451 \ mol \ B$

2. Convert Moles to Atoms

We use Avogadro's Number, 6.02*10²³: the number of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case, the particles are atoms of boron.

$\frac {6.02*10^{23} \ atoms \ B} {1 \ mol \ B}$

Multiply by the number of moles we calculated.

$2.312673451 \ mol \ B *\frac {6.02*10^{23} \ atoms \ B} {1 \ mol \ B}$

The moles of boron cancel.

$2.312673451 *\frac {6.02*10^{23} \ atoms \ B} {1 }$

$2.312673451 *6.02*10^{23} \ atoms \ B} =1.39269195*10^{24} \ atoms \ B$

The original value of grams has 4 significant figures, so our answer should have the same. For the number we calculated, that is the thousandth place.

$1.392\underline69195*10^{24} \ atoms \ B$

The 6 tells us to round the 2 to a 3.

$1.393 *10^{24} \ atoms \ B$

25.00 grams of boron is equal to 1.393*10²⁴ atoms.

6 0

3 years ago