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3 years ago

Need answer for 9.0mm x 2.0mm x 2.0mm

Chemistry

1 answer:

otez555 [7]3 years ago

6 0

(9.0 mm) x (2.0 mm) x (2.0 mm) =

0.036 milliliters

I hope this is correct and helps!

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Costco installs automobile tires on a first-come first-serve basis. the total time a customer needs to wait for the installation

den301095 [7]

There you go that’s the answer

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3 years ago

Hydrogen is manufactured on an industrial scale by this sequence of reactions: Write an equation that gives the overall equilibr

RideAnS [48]

The question is incomplete. The complete question is :

Hydrogen is manufactured on an industrial scale by this sequence of reactions:

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g ) \ \ \ \ \ \ \ \ \ \ K_1$

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \ K_2$

The net reaction is :

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

Write an equation that gives the overall equilibrium constant $K$ in terms of the equilibrium constants $K_1$ and $K_2$ . If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.

Solution :

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g ) \ \ \ \ \ \ \ \ \ \ K_1$

$K_1 = \frac{[CO][H_2]^3}{[CH_4][H_2O]}$ ...............(1)

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \ K_2$

$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$ ...................(2)

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$

On multiplication of equation (1) and (2), we get

$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$

$K_1K_2=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$ .................(4)

Comparing equation (3) and equation (4), we get

$K=K_1K_2$

4 0

2 years ago

If a Hh mom has an offspring with a hh dad, what is the probability that the offspring will have a dominant phenotype?

lisov135 [29]

Answer:

I think 50%

Explanation:

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6 0

2 years ago

Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)

allsm [11]

Answer:

$\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}$

Explanation:

1. Density from mass and volume

$\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}$

2. Volume from density and mass

$V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}$

3. Mass from density and volume

$\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}$

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water = 12.8 mL

Volume of object = 11.8 mL

$\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}$

Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

6 0

3 years ago

[H2O] = 0.077 M

salantis [7]

Cl₂O + H₂O ⇄ 2HClO

K = [HClO]²/[Cl₂O][H₂O]
K = (0,023)²/(0,077×0,077)
K = 0,000529/0,005929
K ≈ 0,0892

:)

7 0

3 years ago