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2 years ago

10

?A scientist is investigating how the lunar cycle affects the number of sea turtles nesting each night. Which of the following i

s the dependent variable?

1 answer:

alexgriva [62]2 years ago

7 0

Answer:

Number of nesting turtles

Explanation:

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What the volume is.

alekssr [168]

The molar mass of NO₂ is 46.0 g/mol

The molar mass of Pb (NO₃)₂ is 331.2 g/mol

First there is a need to find the number of moles of NO₂ via the stoichiometry of reaction:

2Pb(NO₃)₂ → 2PbO (s) + 4NO₂ (g) + 02 (g)

2 × 331.2 g = 4 × 46.0 g

16.87 g = x (mass of NO₂)

mass of NO₂ = 16.87 × 4 × 46 / 2 × 331.2

mass of NO₂ = 3104.08 / 662.4

mass of NO₂ = 4.686 g of NO₂

Now the number of moles are:

1 mole NO₂ = 46.0 g

x moles of NO₂ = 4.686 g

4.686 × 1 / 46.0 = 0.101 moles of NO₂

1 mole = 22.4 L (at STP)

0.101 moles of NO₂ = 0.101 × 22.4 / 1

= 2.26 L

4 0

3 years ago

The most highly concentrated hydrochloric acid is 12 M. What volume of a 12 M stock solution of HCL should be diluted in order t

Reika [66]

The Anderlecht is A. Thank you

8 0

2 years ago

Erica is working in the lab. She wants to remove the fine dust particles suspended in a sample of oil. Which method is she most

BARSIC [14]

Option a is right answer that is reverse osmosis.

7 0

2 years ago

Read 2 more answers

How many moles of AIC3 are there in 1,119.972 g ?

Alchen [17]

Answer:

it is 8.40189 moles of AlCl3

mole = mass/molar mass

mole= 1119.972/133.34

6 0

3 years ago

Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta

shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]$

Putting the values we get :

$\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]$

$\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]$

$\Delta H=-5314.8kJ$

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = $\frac{5314.8}{2}\times 1=2657.4kJ$

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0

2 years ago