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Lorico [155]
3 years ago
5

- 1. Justin usually blows his nose using Kleenex tissues. His snot keeps leaking through the tissue. He decided he is going to c

reate an experiment to find out if there is something better. What is his dependent variable? - A. Justin - B. Amount of snot the tissue can hold - C. Amount of tissue - D. Type of tissue
Chemistry
1 answer:
daser333 [38]3 years ago
5 0

Answer:

B. Amount of snot the tissue can hold

Explanation:

In this specific scenario, the dependent variable would be the amount of snot the tissue can hold. That is because this is what Justin is trying to measure, which is also dependent on the type of tissue which would be the independent variable in this scenario. The different types of tissues can absorb different amounts of snot and this is what Justin ultimately wants to determine.

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The answer would be 1000
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How many gallons are equivalent to 4000.0 ml (3.78L=1gal) show work
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1.0566 gal. 


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6 0
3 years ago
Balance the following reaction. As2S3 + 9O2 → 2As2O3 + SO2
kumpel [21]

Answer:

2As2S3 + 9O2 = 2As2O3 + 6SO2

Explanation:

8 0
3 years ago
Read 2 more answers
Given the following thermodynamic data, calculate the lattice energy of LiCl:
tiny-mole [99]

Answer:

\boxed{\text{-862 kJ/mol}}

Explanation:

One way to calculate the lattice energy is to use Hess's Law.

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?

We must generate this reaction rom the equations given.

(1)  Li(s) + ½Cl₂ (g) ⟶ LiCl(s);      ΔHf°     = -409 kJ·mol⁻¹

(2) Li(s) ⟶ Li(g);                          ΔHsub =    161 kJ·mol⁻¹

(3) Cl₂(g) ⟶ 2Cl(g)                     BE        =   243 kJ·mol⁻¹

(4) Li(g) ⟶Li⁺(g) +e⁻                   IE₁         =   520 kJ·mol⁻¹

(5) Cl(g) + e⁻ ⟶ Cl⁻(g)                EA₁       =  -349 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.

                                                <u>E/kJ </u> 

(5) Li⁺(g) +e⁻ ⟶ Li(g)                520

(6) Li(g) ⟶ Li(s)                         -161

(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s)     -409

(8) Cl(g) ⟶ ½Cl₂(g)                   -121.5

(9) Cl⁻(g) ⟶ Cl(g) + e⁻               <u>+349</u>

      Li⁺(g) +  Cl⁻(g) ⟶ LiCl(s)     -862

The lattice energy of LiCl is \boxed{\textbf{-862 kJ/mol}}.

3 0
3 years ago
Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks
charle [14.2K]

Answer:

pH  = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH-   Ksp = 3X10^-15

Zn+2 + 4OH-   --> Zn(OH)4-2   Kf = 2X10^15

K = Ksp X Kf

  = 3*2*10^-15 * 10^15

  = 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

                Zn(OH)₂ + 2OH⁻(aq)  --> Zn(OH)₄²⁻(aq)

Initial:           0             0.3                      0

Change:                      -2x                     +x

Equilibrium:               0.3 - 2x                 x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²  

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

                                    = 0.3 -(2*0.089)

                                  = 0.122

pOH = -log[OH⁻]

         = -log 0.122

          = 0.91

pH = 14-0.91

     = 13.09

4 0
3 years ago
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