Lewis structures cannot

Consider a balloon that has a volume V. It contains n moles of gas, it has an internal pressure of P, and its temperature is T.

Dmitry [639]

B.) it stays the same

6 0

3 years ago

Read 2 more answers

Explain Charles law

Marianna [84]

The Charles law explains that "volume of an ideal gas at constant pressure is directly proportional to the absolute temperature."

7 0

3 years ago

Calculate the ph of an aqueous solution that is 0.045 m in hcl(aq) at 25°c

ivann1987 [24]

Remember that HCl is a strong acid will dissociate almost 100%
HCl + H₂O --> H₃O⁺ + Cl⁻

pH = -log[H₃O⁺]
pH = -log(0.045) = 1.35

(The rules for sig figs are that if you take the log of a number with x significant figures, then the result should have x significant decimal places.)

The pH is 1.35

4 0

4 years ago

DUPLICATE

Svet_ta [14]

As in relative abundance , one is take reference
So,
One is taken as 1:
Other is subtracted from it:
(1 - 0.6011)(atomic mass of Ga-71)

Equation can be written as:
69.723 = (0.6011)(68.9256) + (1-0.6011)x 
(1-0.6011) is the percentage abundance of Ga-71 expressed in percentage: 

Solving for x 
28.2918 = 0.3989 x 
x= 70.9246.......

7 0

3 years ago

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

5 0

3 years ago