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sergiy2304 [10]
3 years ago
15

Most scientific investigations are driven by the desire to answer a question. true or false

Chemistry
1 answer:
beks73 [17]3 years ago
8 0
This awnser will be true because that's what it is and will always be
You might be interested in
Calculate the molarity of a solution that contains 4.7 moles in 5.2 liters.
AleksAgata [21]

Answer:

0.9M

Explanation:

molarity = mols/L

M=4.7mol/5.2L

= 0.9M

4 0
3 years ago
How many grams of Na2SO4 can be produced from 423.67 g of NaCl?
san4es73 [151]

Mass of  Na2SO4= 514.18 grams

<h3>Further explanation</h3>

Given

423.67 g of NaCl

Required

mass of  Na2SO4

Solution

Reaction

2NaCl + H2SO4 → Na2SO4 + 2HCl

mol NaCl :

= 423.67 g : 58.5 g/mol

= 7.24

From the equation, mol Na2SO4 :

= 1/2 x mol NaCl

= 1/2 x 7.24

= 3.62

Mass  Na2SO4 :

= 3.62 mol x 142,04 g/mol

= 514.18 grams

4 0
3 years ago
Convert 123 in scientific notation
dalvyx [7]
✡ Answer: 1.23*10^2 ✡


- - Add a decimal at the end (to the right) and count till you get to the first number.
So now you have 1.23

- - Now you always want to times it by 10 to the power of how many times you moved it over, in this case, 2

Final answer: 1.23*10^2

✡Hope this helps✡


4 0
3 years ago
Read 2 more answers
Define alkanes and give examples
LenKa [72]
<span>the simplest organic molecules, consisting of only carbon and hydrogen and with only single bonds between carbon atoms.
Ex. Methane, Ethan, Propane, Butane
</span>
3 0
3 years ago
How many excess electrons must be added to an isolated spherical conductor 41.0 cmcm in diameter to produce an electric field of
alina1380 [7]

Answer:

3.65 x 10¹⁰ electrons

Explanation:

we'll apply the following equation for electric field of a point charge on a spherical conductor

E = k \frac{q}{r^{2} }

where E is the electric field

k is a constant of the value 8.99 x 10⁹ Nm²/C²

r is the radius of the spherical conductor

q is the total charge in the sphere

Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)

Electrical field E = 1250 N/C

we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field

E = k \frac{q}{r^{2} }

q = <u>E x r²</u>

        k

q =  <u>1250 N/C x 0.205m</u>²

       8.99 x 10⁹ Nm²/C²

q =   5.84 x 10⁻⁹ C

this is the total charge in the sphere

To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)

n = \frac{q}{e}

n = <u>5.84 x 10⁻⁹ C </u>

       1.6 x 10⁻¹⁹C

n = 3.65 x 10¹⁰ electrons

Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons

3 0
3 years ago
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