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3 years ago

Give the measurement following sig fig rules for tools. Do not forget the unit after your number measurement.

Chemistry

1 answer:

In-s [12.5K]3 years ago

5 0

Answer:

Explanation:

You are not really helped by what is hold the liquid. Beakers and cylinders come on a lot of sizes. I hate to be crabby about things like that, but you really need to be aware that the question is slightly flawed (not your fault).

The beaker, you'd be like to get 1 sig digit. You have to be awfully careful about claiming more. So the and is 50 mL, but that mL is a guess and the 50 is not totally accurate, but what would you say the second digit is? 48 or 47? You don't really know. Maybe even 49.

The graduated cylinder is a little better. Read the bottom of the meniscus (the bottom of the 1/2 bubble). I think you can get 2 sig digs., so the answer is 36 mL. But everything also depends on what you have been told.

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4 years ago

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3 years ago

Consider the balanced reaction: 4Ga(s) + 3O2(g)⟶ 2Ga2O3(s)

Paha777 [63]

First we need to find the number of moles that 43.9g of gallium metal is. We can do this by finding the molar weight of gallium and cross-multiplying to cancel out units:

$\frac{1mole}{69.723g}*\frac{43.9g}{1}=0.63mol_{Ga}$

So we are dealing with 0.63 moles of gallium metal.

We can take from the balanced equation that 4 moles of gallium metal will react completely with 3 moles of oxygen gas. We can take this ratio and make a proportion to find the amount of oxygen gas, in moles, that will react completely with 0.63 moles of gallium metal:

$\frac{4moles_{Ga}}{3moles_{O_{2}}} =\frac{0.63moles_{Ga}}{xmoles_{O_{2}}}$

Cross multiply and solve for x:

$4x=1.89$

$x=0.47moles_{O{2}}$

So now we know that 0.47 moles of oxygen gas will react with 43.9g of gallium metal.

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3 years ago

Give the measurement following sig fig rules for tools. Do not forget the unit after your number measurement.​

Give the measurement following sig fig rules for tools. Do not forget the unit after your number measurement.