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Ghella [55]
4 years ago
15

Which statement best demonstrates how data from a global positioning system (GPS) can be used to lessen the effects of a

Chemistry
1 answer:
dolphi86 [110]4 years ago
3 0

Answer: here is your answer

Explanation: You are visiting your Grandmother and notice that she is eating a balanced diet, taking vitamins, getting the proper amount of sleep and is not overweight. Despite her healthy lifestyle, she appears run down and tired. You realize that it's due to her lack of physical activity. Write a convincing letter to your grandma explaining the benefits of participating in regular physical activity.

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B.

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3 0
3 years ago
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A Bunsen burner's job is to do what?
Vsevolod [243]

Answer:

A Bunsen bumper,named after Robert Bunsen, is a kind of gas burner used as laboratory equipment;it produces a single open gas flame ,and used for heating ,sterilization,and combustion. The gas can be natural gas (which is mainly methane) or a liquefied petroleum gas,such as propane,butane,or a mixture

Explanation:

7 0
4 years ago
What part of the scientific method involves controlling variables while testing a hypothesis? (2 points) Select one: a. Analyzin
scZoUnD [109]

Answer:

D

Explanation:

best choice, makes the most sense

6 0
2 years ago
What is not a mass of 1 mole of a substance?
Alenkinab [10]

Explanation:

been defined in such a way. It is defined as the numbers of particles in 12 g of 12C. If it were 24 g instead of 12 g, then the weight of 1 mole of substance would equal 2 times the atomic/molecular mass in grams.

5 0
3 years ago
Calcium carbonate decomposes at high temperatures to give calcium oxide and carbon dioxide as shown below. CaCO3(s) CaO(s) + CO2
romanna [79]

Answer:

\boxed{\text{10.0 g}}

Explanation:

The balanced equation is

CaCO₃(s) ⇌ CaO(s) + CO₂(g); Kp = 1.16

Calculate Kc

T= 800 °C = 1073 K; Δn= 1

K_{\text{p}} = K_{\text{c}}{RT}^{\Delta n}\\\\1.1 = K_{c}{(RT)}^{1}\\\\K_{c} = \dfrac{1.1}{RT} = \dfrac{1.1}{8.314\times 1073}= \dfrac{1.1}{8921} = 1.233 \times 10^{-4}

Equilibrium concentration of CO₂

K_{c} = [\text{CO}_{2}] = 1.233 \times 10^{-4}

Moles of CO₂ formed

n= \text{5.00 L } \times \dfrac{\text{1.233 $\times$ 10$^{-4}$ \text{ mol} }}{\text{1 L}} = 6.155 \times 10^{-4}\text{ mol}

Moles of CaCO₃ used up

Moles of CaCO₃ used up = moles of CO₂ formed = 6.155 × 10⁻⁴ mol

Mass of CaCO₃ used up

m = 6.155 \times 10^{-4}\text{ mol } \times \dfrac{\text{100.09 g}}{\text{1 mol}} = \text{0.0616 g}

Moles of CaO formed

Moles of CaO formed = moles of CO₂ formed = 6.155 × 10⁻⁴ mol

Mass of CaO formed

m = 6.155 \times 10^{-4}\text{ mol } \times \dfrac{\text{56.08 g}}{\text{1 mol}} = \text{0.0345 g}

Mass of solid at equilibrium

m = 10.0 g – 0.0616 g + 0.0345 g = 10.0 g

\text{The mass of solid remaining at equilibrium is } \boxed{\textbf{10.0 g}}

6 0
3 years ago
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