There are 19.5 g Na in 71.4 g NaHCO₃
Calculate the <em>molecular mass of NaHCO₃</em>.
1 Na = 1 × 22.99 u = 22.99 u
1 H = 1 × 1.008 u = 1.008 u
1 C = 1 × 12.01 u = 12.01 u
3 O = 3 × 16.00 u = <u>48.00 u
</u>
TOTAL = 84.008 u
So, there are 22.99 g of Na in 84.008 g NaHCO₃.
∴ Mass of Na = 71.4 g NaHCO₃ × (22.99 g Na/84.008 g NaHCO₃) = 19.5 g Na
Answer:
C Atomic number
Explanation:
The question is typo it should be 19 is the____ of pottasium. The atomic number is the number on top
Answer:
1.09 × 10⁻⁷ m
UV region
Explanation:
Step 1: Given and required data
Energy of the photon of light (E): 1.83 × 10⁻¹⁸ J
Planck's constant (h): 6.63 × 10⁻³⁴ J.s
Speed of light (c): 3.00 × 10⁸ m/s
Step 2: Calculate the wavelength (λ) of this photon of light
We will use the Planck-Einstein's relation.
E = h × c/λ
λ = h × c/E
λ = 6.63 × 10⁻³⁴ J.s × (3.00 × 10⁸ m/s)/1.83 × 10⁻¹⁸ J = 1.09 × 10⁻⁷ m
This wavelenght falls in the UV region of the electromagnetic spectrum.
question 1
moles = mass/molar mass of Al(OH)3
convert Kg to g
that is 1.09 x 1000=1090g
moles is therefore=1090g/78(molar mass of Al(OH)3)= 13.974 moles
question 2
moles=2.55g/327.2(molar mass of Pb(CO3)2= 7.79 x 10^-3 moles
from avogadro constant
1moles=6.02 x10^23 formula units
what about 7.79 x 10 ^-3
={(7.79 x 10^-3)moles x ( 6.02 x10^23)} /1 mole=4.69 x10^21 formula units
Answer:
copper I oxide
Explanation:
Copper is known to form two oxides; copper I oxide and copper II oxide.
In the question we have the ions; Cu1+ and O2-. Ionic compounds are formed by a combination of ions as typified below;
Cu^+ + O^2- -------> Cu2O
This compound is copper I oxide