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Tatiana [17]
4 years ago
15

An object of mass 5 kilograms is moving across a surface in a straight line with a speed of 3.5 meters/second. What amount of fo

rce would be required to keep the object in motion if you ignore the sureface friction?
Chemistry
1 answer:
vampirchik [111]4 years ago
4 0

F = MA A = change in velocity Constant velocity = no change in velocity = Zero acceleration Ignoring friction, 5 kg * 0 m/s^2 = 0 N Therefore, there is no force (0 N) required to keep the object in motion if the surface friction is ignored.

I hope my answer has come to your help.


Read more on Brainly.com - brainly.com/question/2005686#readmore

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When 10.0 grams of CH4 reacts completely with 40.0 grams of O2 such that there are no reactants left over, 27.5 grams of carbon
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Answer:

\boxed{\text{27.4 g CO$_{2}$; 22.5 g H$_{2}$O}}}

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:      16.04    32.00  44.01   18.02

             CH₄  +  2O₂  →  CO₂ + 2H₂O

m/g:      10.0      40.0

1. Moles of CH₄

\text{Moles of CH}_{4} = \text{10.0 g CH}_{4} \times \dfrac{\text{1 mol CH}_{4}}{\text{16.04 g CH}_{4}} = \text{0.6234 mol CH}_{4}

2. Mass of CO₂

(i) Calculate the moles of CO₂

The molar ratio is (1 mol CO₂ /1 mol CH₄)

\text{Moles of CO$_{2}$} = \text{0.6234 mol CH$_4$} \times \dfrac{\text{1 mol CO$_{2}$}} {\text{1 mol CH$_{4}$}} = \text{0.6234 mol CO$_{2}$}

(ii) Calculate the mass of CO₂

\text{Mass of CO$_{2}$} = \text{0.6234 mol CO$_{2}$} \times \dfrac{\text{44.01 g CO$_{2}$}}{\text{1 mol CO$_{2}$}} = \textbf{27.4 g CO$_{2}$}\\\\\text{The mass of carbon dioxide formed is } \boxed{\textbf{27.4 g CO$_{2}$}}

3. Mass of H₂O

(i) Calculate the moles of H₂O

The molar ratio is (2 mol H₂O /1 mol CH₄)

\text{Moles of H$_{2}$O}= \text{0.6234 mol CH}_{4} \times \dfrac{\text{2 mol H$_{2}$O}}{\text{1 mol CH$_{4}$}} = \text{1.247 mol H$_{2}$O}

(ii) Calculate the mass of H₂O

\text{Mass of H$_{2}$O} = \text{1.247 mol H$_{2}$O } \times \dfrac{\text{18.02 g H$_{2}$O}}{\text{1 mol H$_{2}$O}} = \textbf{22.5 g H$_{2}$O}\\\\\text{The mass of water formed is } \boxed{\textbf{22.5 g H$_{2}$O}}

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