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nalin [4]
3 years ago
9

A 425.00-gram sample of a compound decomposes into 196.01 grams of carbon, 41.14 grams of hydrogen, 130.56 grams of oxygen, and

57.29 grams of silicon. Experiments have shown the compound has a molecular weight of 208.329. What is the molecular formula?
A.
C8H5O2Si
B.
C2H5OSi
C.
C4H10O2Si
D.
C8H20O4Si
Chemistry
1 answer:
Alenkasestr [34]3 years ago
3 0

The molecular formula is D. C_8H_20O_4Si.

<em>Step 1</em>.Calculate the <em>empirical formula </em>

a) Calculate the moles of each element

Moles of C= 196.01 g C × (1 mol C/12.01 g C) = 16.325 mol C  

Moles of H = 41.14 g H × (1 mol H/1.008 g H) = 40.813 mol H

Moles of O = 130.56 g O × (1 mol O/16.00 g O) = 8.1650 mol O

Moles of Si = 57.29 g Si × (1 mol Si/28.085 g Si) = 2.0399 mol Si

b) Calculate the molar ratio of each element

Divide each number by the smallest number of moles and round off to an integer

C:H:O:Si = 8.0027:20.008:4.0027:1 ≈ 8:20:4:1

c) Write the empirical formula

EF = C_8H_20O_4Si

<em>Step </em>2. Calculate the <em>molecular formula</em>

EF Mass = 208.33 u

MF mass = 208.329  u

MF = (EF)_n

n = MF Mass/EF Mass = 208.329 u/208.33 u = 1.0000 ≈ 1

MF = C_8H_20O_4Si

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hope this helps!

6 0
3 years ago
Consider an electron with a mass of 9.11 x 1051 kg and a 100.0 g tennis ball that are both moving with a velocity of 70.0 m s1.
MrRissso [65]

Answer:

(a) 6.38 × 10⁻²⁹ kg·m·s⁻¹; (b) 7.00 kg·m·s⁻¹; (c) 82.7 µm; (d) 7.53 × 10⁻³⁴  m;

(e) Δx ∝ 1/m

Explanation:

(a) Momentum of electron

p = mv = 9.11  × 10⁻³¹ kg × 70.0 m·s⁻¹ = 6.38 × 10⁻²⁹ kg·m·s⁻¹

(b) Momentum of tennis ball

p = mv = 0.1000 kg × 70.0 m·s⁻¹ = 7.00 kg·m·s⁻¹

(c) Δx for electron

Δp = 0.010p = 0.010 × 6.38 × 10⁻²⁹ kg·m·s⁻¹ = 6.38 × 10⁻³¹ kg·m·s⁻¹

\begin{array}{rcl}\Delta x \Delta p & \geq & \dfrac{h}{4 \pi}\\\\\Delta x \times 6.38 \times 10^{-31} \text{ kg$\cdot$m$\cdot$s$^{-1}$} & \geq & \dfrac{6.626 \times 10^{-34} \text{ kg$\cdot$m$^{2}$s}^{-1}}{4 \pi}\\\\\Delta x \times 6.38 \times 10^{-31} & \geq & 5.273 \times 10^{-35} \text{ m}\\\Delta x & \geq & \dfrac{5.273 \times 10^{-35} \text{ m}}{6.38 \times 10^{-31}}\\\\ & \geq&8.27 \times10^{-5} \text{ m}\\ &\geq&\textbf{82.7 $\mu$m}\\\end{array}

(d) Δx for tennis ball

Δp = 0.010p = 0.010 × 7.00 kg·m·s⁻¹ = 0.0700 kg·m·s⁻¹

\begin{array}{rcl}\Delta x \times 0.0700 & \geq & 5.273 \times 10^{-35} \text{ m}\\\Delta x & \geq & \dfrac{5.273 \times 10^{-35} \text{ m}}{0.0700}\\\\ &\geq& \textbf{7.53 $\mathbf{\times 10^{-34}}$ m}\\\end{array}

(e) Relative uncertainty

Both particles are travelling at the same speed, so,

ΔxΔp = Δx × mv = mvΔx = constant

v is constant, so  

Δx ∝ 1/m

Thus, the larger the mass of an object, the smaller the uncertainty in its velocity.

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3 0
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Answer:

B

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6 0
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Amount of extra solute dissolved = 8.97 - 4.29 = 4.7 g

3 0
3 years ago
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