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Ann [662]
3 years ago
7

A/an _______ is a type of research that involves observation and collecting data but does not include a control. A. independent

variable B. experimentation C. dependent variable D. investigation
Chemistry
1 answer:
pantera1 [17]3 years ago
8 0
Should be D. an investigation
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Helppp plss
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Earth's atmosphere contains a huge pool of nitrogen gas (N2). But this nitrogen is “unavailable” to plants, because the gaseous form cannot be used directly by plants without undergoing a transformation. To be used by plants, the N2 must be transformed through a process called nitrogen fixation.

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2 years ago
In an electron dot diagram of propane (C3H8), how many double bonds are present?
NeX [460]

Answer:- none

Explanations:- For naming hydrocarbons we use the suffix -ane, -ene and -yne.

-ane is used when we have only single bonds between all carbons. -ene is used if there is any double bond between two carbons and the -yne is used if there is any triple bond between two carbons.

The given name of the compound is propane. It ends at -ane and so it's an alkane and must have single bonds between all the carbons it has. So, there are zero double bonds present in C_3H_8 .

This is also clear from the below lewis dot structure of the compound.


4 0
3 years ago
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Bakit mahalaga ang paghahanda ng pambansang badyet sa isang taon?​
Andreas93 [3]

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8 0
2 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
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