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fiasKO [112]
3 years ago
14

If the mass of a certain block is 500g, what is the weight of the block?

Chemistry
1 answer:
Dafna11 [192]3 years ago
5 0

<em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em>

<em>Sol</em><em>ution</em><em>,</em>

<em>Mass</em><em>=</em><em>5</em><em>0</em><em>0</em><em>g</em><em> </em><em>=</em><em>5</em><em>0</em><em>0</em><em>/</em><em>1000</em><em>=</em><em>0</em><em>.</em><em>5</em><em> </em><em>kg</em>

<em>gravity</em><em>(</em><em>g</em><em>)</em><em>=</em><em>9</em><em>.</em><em>8</em><em>m</em><em>/</em><em>s^</em><em>2</em>

<em>Now</em><em>,</em>

<em>Weight</em><em>=</em><em> </em><em>m</em><em>*</em><em>g</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>0</em><em>.</em><em>5</em><em>*</em><em>9</em><em>.</em><em>8</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>4</em><em>.</em><em>9</em><em> </em><em>N</em>

<em>So</em><em> </em><em>the</em><em> </em><em>weight</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>block</em><em> </em><em>is</em><em> </em><em>4</em><em>.</em><em>9</em><em> </em><em>Newton</em>

<em>hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em><em>.</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>

<em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em>

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The chemical equation representing the reaction of silver nitrate with barium chloride:

2AgNO_{3}(aq) + BaCl_{2}(aq)--> 2AgCl (s) + Ba(NO_{3})_{2}(aq)

Given mass of barium chloride = 4.62 g

Moles of BaCl_{2} = 4.62 g BaCl_{2}*\frac{1 mol BaCl_{2}}{208.23 g BaCl_{2}} =   0.0222 mol BaCl_{2}

Moles of AgCl = 0.0222 mol BaCl_{2} * \frac{2 mol AgCl}{1 mol BaCl_{2}} = 0.0444 mol AgCl

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<em></em>

I hope it helps!

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