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inessss [21]
2 years ago
12

What is the chemical name of N₂O

Chemistry
1 answer:
GrogVix [38]2 years ago
5 0

Nitrous oxide .

The Lewis dot structure is attached

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List the sub-atomic particles in the order in which they were discovered and tell who discovered each one.
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<span>Photons were the first sub-atomic particles detected, but not quite discovered as they could not be explained. Photons were first detected by Johann Wilhelm Ritter, Victor Schumann, and Winhelm Rontgen. The next, and first sub-atomic particle discovered, was the electron. The electron was discovered by J. J. Thompson in the late 1800s. The next two sub-atomic particle discoveries were the alpha particle and photon, discovered by Ernest Rutherford and Paul Villard respectively. Rutherford also discovered the proton and in 1932, James Chadwick discovered the neutron.</span>
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3 years ago
How do you balance the following equation only using coefficients?
yKpoI14uk [10]

The answer will be 6CO2 + 6H2O ——-> 1 C6H12O6 + 6O2

3 0
3 years ago
Monel metal is a corrosion-resistant copper-nickel alloy used in the electronics industry. A particular alloy with a density of
klemol [59]
<span>To find the volume of the plate without accounting for the hole firstly
V = (15.0 cm)(12.5 cm)(0.250 cm) = 46.875 cm^3
and the volume of the hole is
(pi)(1.25 cm)^2(0.250 cm) = 1.2272 cm^3
we will subtract the volume of the hole from the rest 45.648 cm^3
the multiply this by the density of the alloy to find the mass
(8.80 g/cm^3)(45.648 cm^3) = 401.701 g.
0.044% of this is Si, so (0.00044)(401.701 g) = 0.17675 g is silicon.
by the number of atoms and using average atomic mass of silicon and Avogadro's number to find the number of silicon atoms:
(0.17675 g)(1 mol/28.0855 g)(6.022E23 atoms/1 mol) =3.794E21atoms of Si
3.10% of these are Si-30:(0.0310)(3.794E18 atoms)=1.176E20 atoms of Si-30 and with two significant figures, 1.2E20 atoms.
hope this helps
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4 0
3 years ago
Atom , classify it<br><br><br><br><br>helppoppppoeowiie​
timama [110]

Answer:

A atom is a small unit to which matter can be divided without releasing electrically charged particles

Explanation:

5 0
3 years ago
An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for
8090 [49]

Answer: The freezing point and boiling point of the solution are -6.6^0C and 101.8^0C respectively.

Explanation:

Depression in freezing point:

T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_f = freezing point of solution = ?

T^o_f = freezing point of water = 0^0C

k_f = freezing point constant of water = 1.86^0C/m

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

w_2 = mass of solute (ethylene glycol) = 21.4 g

w_1= mass of solvent (water) = density\times volume=1.00g/ml\times 97.6ml=97.6g

M_2 = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}

T_f=-6.6^0C

Therefore,the freezing point of the solution is -6.6^0C

Elevation in boiling point :

T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of water = 100^0C

k_b = boiling point constant of water = 0.52^0C/m

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

w_2 = mass of solute (ethylene glycol) = 21.4 g

w_1= mass of solvent (water) = density\times volume=1.00g/ml\times 97.6ml=97.6g

M_2 = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}

T_b=101.8^0C

Thus the boiling point of the solution is 101.8^0C

4 0
3 years ago
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