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natta225 [31]
3 years ago
10

A chemist has a block of aluminum metal (density is 2.7 g/mL). The block weighs 1.5. What is the volume of the aluminum block?

Chemistry
1 answer:
12345 [234]3 years ago
4 0

Answer:

0.56 mL

Explanation:

Volume = mass ÷ density

Volume = 1.5 ÷ 2.7 g/mL

Volume = 0.5555555556 = 0.56 mL

The volume of the aluminum block is 0.56 mL.

Hope this helps. :)

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you are holding a bowling ball with a mass of 6 kg at a height of 1 meter. How much gravitational potential energy does the bowl
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The formula for GPE is PE=mgh, where “m” is the mass of the object, “g” is the acceleration due to gravity (~9.8 m/s^2 on Earth’s surface), and “h” is the height of the object from the ground. Therefore,

PE=mgh

PE=(6 kg)(9.8 m/s^2)(1 m)

PE=58.8 kg•m^2/s^2 or 58.8 Newtons

The GPE of the bowling ball under these conditions would be about 59 Newtons.

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# Cary on learning

8 0
2 years ago
What is the color of benzene and bromine​
Harman [31]
Bromine is a reddish-brown color.
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6 0
3 years ago
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The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentrat
Valentin [98]

Answer:

(a)

0.0342M

(b)

t_{1/2}=17.36s\\t_{1/2}=23.15s

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\

[NOBr]=\frac{1}{29.2/M}=0.0342M

(b) Now, for a second-order reaction, the half-life is computed as shown below:

t_{1/2}=\frac{1}{k[NOBr]_0}

Therefore, for the given initial concentrations one obtains:

t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.072M}=17.36s\\t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.054M}=23.15s

Best regards.

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2 years ago
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