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2 years ago

Answer the question please

Chemistry

2 answers:

RUDIKE [14]2 years ago

7 0

Answer:

129 degrees

Explanation:

100 - (-29) = 129

NemiM [27]2 years ago

3 0

City A: 100° F
City B: -29° F

100 - (-29) ==> 100 + 29 = 129 degrees difference.

The different between City A and B is 129° F

Hope this helps!

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Two substances that have the same number of moles of representative particles always have: Answer the same number of atoms the s

Nesterboy [21]

<span>same no of atoms
b/c no. of atoms = moles * avogadro's no (6.022 X 10^23)</span>

7 0

3 years ago

Read 2 more answers

5.5x10² – 6.50x10'<br> Express the calculation in correct significant digits

OLga [1]

Answer:

$6.15x10^2$

Explanation:

Hello!

In this case, when performing mathematical operations with numbers in scientific notation, the first step is to write them in standard notation:

550 - 65.0

Thus, the result without any significant figures-based analysis is:

615.0

However, since 550 is significant to the ones and 65.0 to the tenths, we need to round the result to the bigger significance, in this case to the ones; therefore the appropriate result would be:

615

That in scientific notation would be:

$6.15x10^2$

Best regards!

5 0

3 years ago

A large gently sloping volcano with a fluid eruption

kozerog [31]

Salutations!

A large gently sloping volcano with a fluid eruption is called shield volcano. The fluid is basaltic lava and the slops are gently sliding.

Hope I helped (:

Have a great day!

6 0

3 years ago

A diprotic acid, H₂A, has Ka1 = 3.4 × 10⁻⁴ and Ka2 = 6.7 × 10⁻⁹. What is the pH of a 0.18 M solution of H₂A?

Tcecarenko [31]

Answer:

pH = 2.10

Explanation:

We name an acid as diprotic because it can release two protons:

H₂A + H₂O ⇄ H₃O⁺ + HA⁻ Ka₁

HA⁻ + H₂O ⇄ H₃O⁺ + A⁻² Ka₂

We propose the mass balance:

Analytical concentration = [H₂A] + [HA⁻] + [A⁻²]

As Ka₂ is so small, we avoid the [A⁻²] so:

0.18 M = [H₂A] + [HA⁻]

But we can not avoid the HA⁻, because the Ka₁. Ka₁'s expression is:

Ka₁ = [H₃O⁺] . [HA⁻] / [H₂A]

We propose the charge balance:

[H₃O⁺] = [HA⁻] + [A⁻²] + [OH⁻]

As we did not consider the A⁻², we can miss the term and if

Kw = H⁺ . OH⁻

We replace Kw/H⁺ = OH⁻. So the new equation is:

[H₃O⁺] = [HA⁻] + Kw / [H₃O⁺]

The acid is so concentrated, so we can avoid the term with the Kw, so:

[H₃O⁺] = [HA⁻]

In the mass balance we would have:

0.18 M = [H₂A]

We replace at Ka₁

Ka₁ = [H₃O⁺] . [HA⁻] / [H₂A]

Ka1 . 0.18 / [H₃O⁺] = [HA⁻]

We replace at the charge balance:

[H₃O⁺] = Ka1 . 0.18 / [H₃O⁺]

[H₃O⁺]² = 3.4×10⁻⁴ . 0.18

[H₃O⁺] = √(3.4×10⁻⁴ . 0.18)

[H₃O⁺] = 7.82×10⁻³

- log [H₃O⁺] = pH → - log 7.82×10⁻³

pH = 2.10

5 0

4 years ago

A gas that has a volume of 28 liters, a temperature of 42°C, and an unknown pressure has its volume increased to 49 liters and i

kicyunya [14]

Answer:

7.35 atm

Explanation:

From the question we are given;

Initial volume, V1 as 28 L
Initial temperature, T1 as 42°C or 315 K
New volume. V2 as 49 L
New temperature, T2 as 27°C or 300 K
New Pressure, P2 as 4.0 atm

We are required to determine the initial pressure, P1

We are going to use the combined gas law...

According to the combined gas law,

$\frac{P_{1} V_{1} }{T_1} =\frac{P_2V_2}{T_2}$

Therefore, to get P1

$P_1=\frac{P_2V_2T_1}{V_1T_2}$

We get;

$P_1=\frac{(4.0)((49)(315)}{(300)(28)}$

$P_1=7.35$

Therefore, the initial pressure of the gas is 7.35 atm

4 0

3 years ago