Answer:
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Explanation:
Its C
a catalyst speeds up a reaction by offering the reaction an alternative reaction pathway with a lower activation energy
hope that helps
Hey there!
Ca + H₃PO₄ → Ca₃(PO₄)₂ + H₂
Balance PO₄.
1 on the left, 2 on the right. Add a coefficient of 2 in front of H₃PO₄.
Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + H₂
Balance H.
6 on the left, 2 on the right. Add a coefficient of 3 in front of H₂.
Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂
Balance Ca.
1 on the right, 3 on the right. Add a coefficient of 3 in front of Ca.
3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂
Our final balanced equation:
3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂
Hope this helps!
Answer:

Explanation:
Hello there!
In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

In terms of mass, specific heat and temperature change is:

Now, solve for the final temperature, as follows:

Then, plug in the masses, specific heat and temperatures to obtain:

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.
Regards!
Use the rule of 3 simple
50 g mean 100% 0,1895 mean x% cross multiplie and calcule the x value x = 0,1895*100/50 = ?%