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Paraphin [41]
3 years ago
11

What is four thing the atmosphere does for us

Chemistry
2 answers:
pogonyaev3 years ago
5 0
Proveds us air to breath and keeps the hot air out sory but only got 2
artcher [175]3 years ago
4 0
Keeps in heat, protects us from radiation, provides oxygen, and protects us from objects coming towards the earth.
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The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C
Kipish [7]

Answer: 72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Putting in the values we get:

\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

According to stoichiometry:

1 mole of C_3H_8 requires 5 moles of oxygen

0.45 moles of C_3H_8 require= \frac{5}{1}\times 0.45=2.25 moles of oxygen

Mass of O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g

72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

7 0
3 years ago
• What is produced at the end of the cell cycle? How do they compare to each other and
olga2289 [7]

Answer:

At the end of cell cycle (mitosis replication), two daughter cells are produced. Each contains exactly the same number of DNA content.

6 0
3 years ago
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
Please help I will reward brainly
Hoochie [10]
The answer is A, lunar.
3 0
3 years ago
Zsdfgxcghjhvkbjlnm;',
Zielflug [23.3K]
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8 0
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