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Sonja [21]
3 years ago
6

The volume of water in a graduated cylinder was increased from 52.0 mL to 75.5 mL when a piece of irregular metal was placed in

the water. If the piece of metal has a mass of 265.6 g, what is the density of the metal?
Chemistry
2 answers:
Artyom0805 [142]3 years ago
6 0

Answer:

<h2>11.30 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

volume = final volume of water - initial volume of water

volume = 75.5 - 52 = 23.5

From the question we have

density =  \frac{265.6}{23.5}  \\  = 11.302127...

We have the final answer as

<h3>11.30 g/mL</h3>

Hope this helps you

olga nikolaevna [1]3 years ago
5 0

Answer:

Approximately 11.3 g/mL

Explanation:

The formula for density is mass divided by density. In this problem, the mass is already given: 265.6 grams. The density is the part with the milliliters. 75.5 mL - 52.0 mL is 23.5 mL.

After plugging these numbers into the formula, you get 265.6 g/23.5 mL. After divided those numbers, you get 11.30212..... g/mL, which rounds to 11.30 g/mL.

I hope this helps!

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The following data were measured for the reaction BF3(g)+NH3(g)→F3BNH3(g): Experiment [BF3](M) [NH3](M) Initial Rate (M/s) 1 0.2
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Answer:

-r_{A}=k\times[BF_3]^{1}\times[NH_3]^{1}

Explanation:

The rate law of a chemical reaction is given by

-r_{A}=k\times[BF_3]^{\alpha}\times[NH_3]^{\beta}

This law can be written for any experiment, and making the quotient between those expressions the reaction orders can be found

Between experiments 1 and 2  

\frac{-r_{A1}}{{-r}_{A2}}=\left(\frac{\left[NH_3\right]_1}{\left[NH_3\right]_2}\right)^\beta

Then the expression for the calculation of \beta

\beta=\frac{ln\frac{-r_{A1}}{-r_{A2}}}{ln\left(\frac{\left[NH_3\right]_1}{\left[NH_3\right]_2}\right)}=\frac{ln\frac{0.2130}{0.1065}}{ln\left(\frac{0.250}{0.125}\right)}

Resolving  

\beta=1

Doing the same between experiments 3 and 4 the expression for \alpha is

\alpha=\frac{ln\frac{-r_{A3}}{-r_{A4}}}{ln\left(\frac{\left[BF_3\right]_3}{\left[BF_3\right]_4}\right)}=\frac{ln\frac{0.0682}{0.1193}}{ln\left(\frac{0.200}{0.350}\right)}

Resolving  

\alpha=1

This means that the rate law for this reaction is  

-r_{A}=k\times[BF_3]^{1}\times[NH_3]^{1}

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