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3 years ago

5

Light shining on a strip of metal can dislodge electrons. Do you think this is more consistent with light being made up of waves

or of particles?

2 answers:

maks197457 [2]3 years ago

8 0

particles is the right answer.

Gnoma [55]3 years ago

7 0

Light would be more consistent being made up of particles rather than waves.

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Does Ag metal reacts with NaOH

joja [24]

No Ag cannot react with NaOH because Ag is less reactive than Na in the reactivity series and can't displace it

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3 years ago

How many federal agencies are responsible for managing land resources in the United States?

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The answer would be 4.

The agencies are:

Department of Agriculture's Forest Service (AFS)

Department of the Interior's Bureau of Land Management (BLM)

Fish and Wildlife Service (FWS)

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I learned about these agencies in school.

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3 years ago

How many moles in 30.0 grams of h3po4

jok3333 [9.3K]

Molar mass H₃PO₄ = 98.0 g/mol

1 mole ----- 98.0 g
? mole ------ 30.0 g

moles = 30.0 * 1 / 98.0

= 0.306 moles

hope this helps!

5 0

3 years ago

Which of the following equations has the correct products and is balanced correctly for a reaction between Na3PO4 and KOH?

ASHA 777 [7]

The answer is: A) Na3PO4 + 3KOH → 3NaOH + K3PO4, because K retains the same charge throughout the reaction.

This chemical reaction is double displacement reaction - cations (K⁺ and Na⁺) and anions (PO₄³⁻⁻ and OH⁻) of the two reactants switch places and form two new compounds.

Na₃PO₄ is sodium phosphate.

KOH is potassium hydroxide.

NaOH is sodium hydroxide.

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According to the mass conservation law, there are same number of atoms on both side of balanced chemical reaction.

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3 years ago

What is the ratio of [a–]/[ha] at ph 2.75? the pka of formic acid (methanoic acid, h–cooh) is 3.75?

Varvara68 [4.7K]

The dissociation of formic acid is:

$HCOOH \rightleftharpoons HCOO^{-} + H^{+}$

The acid dissociation constant of formic acid, $k_a$ is:

$k_a = \frac{[HCOO^{-}] [H^{+}]}{HCOOH}$

Rearranging the equation:

$\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}$

pH = 2.75

$pH = -log[H^{+}]$

$[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}$

$pk_a = 3.75$

$k_a = 10^{-3.75} = 1.78\times 10^{-4}$

Substituting the values in the equation:

$\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}$

$\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}$

Hence, the ratio is $\frac{1}{10}$ .

4 0

3 years ago