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3 years ago

Why would you weight less on a high mountain peak than you would at sea level?

Chemistry

1 answer:

shutvik [7]3 years ago

3 0

Answer:

because weight depends on the gravity, gravity decrease s with increasing altitude,hence I have less weight

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What part is the independent variable and what part is the dependent variable in the scenario: The blood pressure of a soldier i

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What part is the independent variable and what part is the dependent variable of the seminary the blood pressure of a soldier is measured while he’s resting soldiers and exposed to a stressful environment and his blood pressure is measured in

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2 years ago

H is a chemical _____ and H2 is a chemical ______

jeyben [28]

I don’t know if I am right but I think it is symbol for the first one and the second one is formula.

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2 years ago

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A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)↽−−⇀H+(aq)+A−(aq) The equilibrium concentratio

Pachacha [2.7K]

Answer:

$pK_{a}$ of HA is 6.80

Explanation:

$pK_{a}=-logK_{a}$

Acid dissociation constant ( $K_{a}$ ) of HA is represented as-

$K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}$

Where species inside third bracket represents equilibrium concentrations

Now, plug in all the given equilibrium concentration into above equation-

$K_{a}=\frac{(2.00\times 10^{-4})\times (2.00\times 10^{-4})}{0.250}$

So, $K_{a}=1.6\times 10^{-7}$

Hence $pK_{a}=-log(1.6\times 10^{-7})=6.80$

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2 years ago

1. Complete the following

Marat540 [252]

Answer:

$\large \boxed{\text{0.603 mol}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of O₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 18.1

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{18.1 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1005 mol C$_{6}$H$_{12}$O}_{6}$

b) Moles of O₂

$\text{Moles of O}_{2} =\text{0.1005 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol O}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.603 mol O}_{2}\\\\\text{The reaction requires $\large \boxed{\textbf{0.603 mol}}$ of oxygen}$

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2 years ago

Spooky, scary skeletons

konstantin123 [22]

Answer:

I LOVE THIS OMG

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2 years ago

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