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Evgesh-ka [11]
2 years ago
12

Can you write the answer for lot of points ​

Chemistry
2 answers:
ruslelena [56]2 years ago
4 0

\large\orange{\sf{Given:}}

<h3>Work (W)=2400 Joules</h3><h3>Distance(D)=6 metres</h3>

\large\blue{\sf{To\:find:}}

<h3>Force=?</h3>

\large\red{\sf{Solution:}}

  • The formula we are using to solve is:

\begin{gathered}\\ \sf\longmapsto Work=Force×Distance\end{gathered}

\begin{gathered}\\ \sf\longmapsto 2400=Force×6\end{gathered}

\begin{gathered}\\ \sf\longmapsto Force=\frac{2400}{6}\end{gathered}

\begin{gathered}\\ \sf\longmapsto Force=\cancel{\dfrac{2400}{6}}\end{gathered}

\begin{gathered}\\ \sf\longmapsto Force=400N\end{gathered}

Ostrovityanka [42]2 years ago
4 0
  • Work=2400J
  • Displacement=6m

\\ \sf\longmapsto Work=Force\times Displacement

\\ \sf\longmapsto Force=\dfrac{Work}{Displacement}

\\ \sf\longmapsto Force=\dfrac{2400}{6}

\\ \sf\longmapsto Force=400N

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How much ch2o is needed to prepare 445 ml of a 2.65 m solution of ch2o?
worty [1.4K]

Answer: 35.4 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

Molality = 2.65

n= moles of solute =?

 V_s = volume of solution in ml = 445 ml

Putting in the values we get:

2.65=\frac{n\times 1000}{445ml}

n=1.18

Mass of solute in g=moles\times {\text {molar mass}}=1.18mol\times 30.02g/mol=35.4g

Thus 35.4 grams of CH_2O is needed to prepare 445 ml of a 2.65 m solution of CH_2O.

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If we add a catalyst to the following equation, CO + H2O + heat CO2 + H2, which way will the equilibrium shift?
Igoryamba

Answer:

No effect.

Explanation:

Hello,

In this case, considering the widely studied Le Chatelier's principle, we can realize that the factors affecting equilibrium are concentration, temperature and pressure and volume if the reaction is in gaseous phase and with non-zero change in the number of moles. In such a way, by adding a catalyst to given reaction will have no effect on the equilibrium direction.

Best regards.

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