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2 years ago

5

First to answer correctly gets brainleist

1 answer:

LenaWriter [7]2 years ago

3 0

Answer:

c. Waxing crescent

- I hope this helps have a great night

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Using the accepted value for the volume of 1 gram of water at the temperature of the room that you reported above, what is the a

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Write procedural steps that allow you to demonstrate the sun's role in the water cycle using common material - for each explain what you are modeling and how the materials you have chosen represent nature.

Using the accepted value for the volume of 1 gram of water at the temperature of the room that you reported above, what is the accepted value for the density of water

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3 years ago

Which statement best describes how an existing theory is often affected by the development of new technology?

grigory [225]

Answer: An existing theory is modified so that it can explain both the old and new observations.

Explanation:

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3 years ago

The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?

anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

$K_a\times K_b=K_w$

$K_b$ for sodium formate is $K_b=\frac {K_w}{K_a}$

Given that:

$K_a$ of formic acid = $1.8\times 10^{-4}$

And, $K_w=10^{-14}$

So,

$K_b=\frac {10^{-14}}{1.8\times 10^{-4}}$

$K_b=5.5556\times 10^{-11}$

Concentration = 0.35 M

HCOONa ⇒ Na⁺ + HCOO⁻

Consider the ICE take for the formate ion as:

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻

At t=0 0.35 - -

At t =equilibrium (0.35-x) x x

The expression for dissociation constant of sodium formate is:

$K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}$

$5.5556\times 10^{-11}=\frac {x^2}{0.35-x}$

Solving for x, we get:

x = 0.44×10⁻⁵ M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

5 0

3 years ago

If molar mass of M(OH)3 = 78 8. mass of M

Cerrena [4.2K]

Answer:

From molar mass=total RAM of each individual element

78.8=(16+1)×3+M

78.8-51=M

27.8g/mol=M

5 0

3 years ago

At equilibrium at 2500K, [HCl]=0.0625M and [H2]=[Cl2]=0.00450M for the reaction H2+Cl2 ⇌ HCl.

ASHA 777 [7]

Answer:

a. $H_2+Cl_2 \rightleftharpoons 2HCl$

b. K = 192.9

c. Products are favored.

Explanation:

Hello!

a. In this case, according to the unbalanced chemical reaction we need to balance HCl as shown below:

$H_2+Cl_2 \rightleftharpoons 2HCl$

In order to reach 2 hydrogen and chlorine atoms at both sides.

b. Here, given the concentrations at equilibrium and the following equilibrium expression, we have:

$K=\frac{[HCl]^2}{[H_2][Cl_2]}$

Therefore, we plug in the data to obtain:

$K=\frac{(0.0625)^2}{(0.00450)(0.00450)}\\\\K=192.9$

c. Finally, we infer that since K>>1 the forward reaction towards products is favored.

Best regards!

8 0

2 years ago