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3 years ago

True or false climatologist use weather balloon stethoscope and satellite

Chemistry

2 answers:

Anastaziya [24]3 years ago

7 0

False, have a good great amazing day

Alona [7]3 years ago

5 0

False I think because they don't use balloons to check the weather

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Balance the equation in the box. Click in the answer box to activate the palette. N2(g) + H2(g) → NH3(g)

lesantik [10]

Answer:

N2(g) + 3H2(g) → 2 NH3(g)

Explanation:

N2(g) + H2(g) → NH3(g)

We start equaling the number of N atoms in both sides multiplying by 2 the NH3.

N2(g) + H2(g) → 2 NH3(g)

So we equals the H atoms (there are six in products sites)

N2(g) + 3 H2(g) → 2 NH3(g)

7 0

3 years ago

4K + O2 + 2K20<br> Which statement is true?

Jet001 [13]

Answer:

yes 4K + O2 ------>2K20 is true.

5 0

2 years ago

Given pH = 8.55 Find: [H3O+] and [OH-] Is this acidic, basic or neutral?

Firlakuza [10]

Answer:

Explanation:

The given pH = 8.55

Unknown:

[H₃O⁺] = ?

[OH⁻] = ?

In order to find these unknowns we must first establish some relationship.

pH = -log[H₃O⁺]

8.55 = -log[H₃O⁺]

[H₃O⁺] = inverse log₁₀(-8.55) = 2.82 x 10⁻⁹moldm⁻³

To find the [OH⁻],

pH + pOH = 14

pOH = 14 - pH = 14 - 8.55

pOH = 5.45

pOH = -log[OH⁻]

[OH⁻] = inverse log₁₀ (-5.45) = 3.55 x 10⁻⁶moldm⁻³

The solution is basic because it has more concentration of OH⁻ ions compared to H⁺ ions.

7 0

3 years ago

Holly has an unknown substance in a beaker. she wants to determine the relative ph of the unknown substance. she places a piece

AURORKA [14]

Blue litmus paper turns red in the presence of an acid. Therefore, it can be assumed that the substance in the beaker is an acid.

Acids have a pH level of less than 7. Consequently, it can be assumed that the substance has a pH level less than 7.

6 0

3 years ago

Read 2 more answers

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$

Thus it takes $0.0241 \; \text{L}$ of sodium hydroxide to produce this buffer solution.

6 0

3 years ago