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Shtirlitz [24]
3 years ago
12

How many electrons does Br lose or gain to fill its octet?

Chemistry
1 answer:
Greeley [361]3 years ago
4 0

i think it is 8. I might be wrong.

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Describe, on a molecular level, how you would expect these lipids to behave in water.
Gelneren [198K]
Lipids are hydrophobic; They would be insoluble, group together, and float to the top
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Solar, wind, and water power are renewable resources and can replace our dependence on fossil fuels. They each have their own dr
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Solar- the sun is not always shining, for example at night you can't get any energy from the sun
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4 years ago
6. Consider the reaction: CzHo (g) + 02 (8) - 4 CO2(g) + 6H2O (1)
deff fn [24]

Answer:

1120 gm

Explanation:

6. Consider the reaction: CzHo (g) + 02 (8) - 4 CO2(g) + 6H2O (1)

(a) Balance the equation.

(b) How many grams of oxygen are required to react with 10 moles of ethane for a complete

combustion reaction?

FIRST, CORRECT THE EQUATION THEN BALANCE

2C2H6(G) + 7O2------------>  4CO2  + 6H2O

so for 10 moles of ethane, we need

7 X 5 = 35 MOLES O2

=35 MOLES O2

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6 0
3 years ago
What mass of solid lead would displace exactly 234.6 liters of water?
eimsori [14]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

The mass of solid lead would displace exactly 234.6 liters of water should be <span>2,674,440</span>
6 0
3 years ago
Read 2 more answers
At its critical point, ammonia has a density of 0.235 g cm23. You have a special thick­walled glass tube that has a 10.0­mm outs
salantis [7]

Answer:

\large \boxed{\text{69.3 mg}}

Explanation:

1. Volume of sealed tube

Assume the sealed tube is a right circular cylinder in which the cap and the base are also 4.20 mm thick.

Its outside dimensions are 155 mm long × 10.0 mm diameter.

Its inside dimensions are

h = 155 mm - 2 × 4.20 mm = 146.6 mm

r = 5.0 mm - 4.20 mm = 0.8 mm

V = πr²h = π(0.8)²× 146.6 mm³ = 294.8 mm³ = 0.2948 cm³

2. Calculate the mass of NH₃

\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\0.235 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{0.2948 cm}^{3}}\\\\m & = & \text{0.0693 g}\\& = & \textbf{69.3 mg}\\\end{array}\\\text{You must seal $\large \boxed{\textbf{69.3 mg}}$ of ammonia in the tube.}

4 0
3 years ago
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