During convention _______, less dense fluid rises and ______

Please help me I’ll give brainliest

astraxan [27]

Answer: A

Explanation: Boiling point increases with the rise of electrons.

3 0

3 years ago

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You have two cups one is 6cm and other 10cm with no measurement and you have to pour 8cm of water into the 10cm cup using the tw

Korolek [52]

Here’s one way to do it.

1. Fill the 6 cm cup.

2. Pour its contents into the 10 cm cup. This leaves 4 cm yet to be filled.

3. Refill the 6 cm cup and use it to fill the 10 cm cup. This leaves 2 cm in the 6 cm cup.

4. Empty the 10 cm cup and add the 2 cm from the 6 cm cup.

5. Refill the 6 cm cup.

6. Pour its contents into the 10 cm cup.

The 10 cm cup now contains 8 cm of water.

5 0

3 years ago

Name at least 6 methods of seperation

coldgirl [10]

Answer:

Handpicking
Threshing
Winnowing
Sieving
Evaporation
Distillation
Filtration or Sedimentation
Separating Funnel
Magnetic Separation

Explanation:

You said at least 6 methods meaning I can give more than 6.

Hope it helps.

4 0

3 years ago

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Your friend Niko predicts that iron and silver nitrate will form when silver is placed into an iron (III) nitrate solution. Is t

SCORPION-xisa [38]

Answer:

It is correct because iron is more reactive than silver.

Explanation:

3 0

3 years ago

After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108. Expre

Marina CMI [18]

Answer:

$\large \boxed{1.77 \times 10^{-5}\text{ mol/L}}$

Explanation:

Assume that you have mixed 135 mL of 0.0147 mol·L⁻¹ NiCl₂ with 190 mL of 0.250 mol·L⁻¹ NH₃.

1. Moles of Ni²⁺

$n = \text{135 mL} \times \dfrac{\text{0.0147 mmol}}{\text{1 mL}} = \text{1.984 mmol}$

2. Moles of NH₃

$n = \text{190 mL} \times \dfrac{\text{0.250 mmol}}{\text{1 mL}} = \text{47.50 mmol}$

3. Initial concentrations after mixing

(a) Total volume

V = 135 mL + 190 mL = 325 mL

(b) [Ni²⁺]

$c = \dfrac{\text{1.984 mmol}}{\text{325 mL}} = 6.106 \times 10^{-3}\text{ mol/L}$

(c) [NH₃]

$c = \dfrac{\text{47.50 mmol}}{\text{325 mL}} = \text{0.1462 mol/L}$

3. Equilibrium concentration of Ni²⁺

The reaction will reach the same equilibrium whether it approaches from the right or left.

Assume the reaction goes to completion.

Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺

I/mol·L⁻¹: 6.106×10⁻³ 0.1462 0

C/mol·L⁻¹: -6.106×10⁻³ 0.1462-6×6.106×10⁻³ 0

E/mol·L⁻¹: 0 0.1095 6.106×10⁻³

Then we approach equilibrium from the right.

Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺

I/mol·L⁻¹: 0 0.1095 6.106×10⁻³

C/mol·L⁻¹: +x +6x -x

E/mol·L⁻¹: x 0.1095+6x 6.106×10⁻³-x

$K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}$

Kf is large, so x ≪ 6.106×10⁻³. Then

$K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}\\\\\dfrac{6.106 \times 10^{-3}}{x\times 0.1095^{6}} = 2.0 \times 10^{8}\\\\6.106 \times 10^{-3} = 2.0 \times 10^{8}\times 0.1095^{6}x= 345.1x\\x= \dfrac{6.106 \times 10^{-3}}{345.1} = 1.77 \times 10^{-5}\\\\\text{The concentration of Ni$^{2+}$ at equilibrium is $\large \boxed{\mathbf{1.77 \times 10^{-5}}\textbf{ mol/L}}$}$

3 0

3 years ago

During convention _______, less dense fluid rises and _______, denser fluid sinks

During convention _, less dense fluid rises and _, denser fluid sinks