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3 years ago

11. Sound's loudness (or volume) is measured in decisels, A True B. False

Chemistry

1 answer:

bagirrra123 [75]3 years ago

7 0

Answer:

sound loudness is measured in decibels, A. True

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Americium-241 is used in smoke detectors. it has a first order rate constant for radioactive decay of k=1.6×10−3yr−1. by contras

Helga [31]

This question is missing the part that actually asks the question. The questions that are asked are as follows:

(a) How much of a 1.00 mg sample of americium remains after 4 day? Express your answer using 2 significant figures.

(b) How much of a 1.00 mg sample of iodine remains after 4 days? Express your answer using 3 significant figures.

We can use the equation for a first order rate law to find the amount of material remaining after 4 days:

[A] = [A]₀e^(-kt)

[A]₀ = initial amount
k = rate constant
t = time
[A] = amount of material at time, t.

(a) For americium we begin with 1.00 mg of sample and must convert time to units of years, as our rate constant, k, is in units of yr⁻¹.

4 days x 1 year/365 days = 0.0110

A = (1.00)e^((-1.6x10^-3)(0.0110))
A = 1.0 mg

The decay of americium is so slow that no noticeable change occurs over 4 days.

(b) We can simply plug in the information of iodine-125 and solve for A:

A = (1.00)e^(-0.011 x 4)
A = 0.957 mg

Iodine-125 decays at a much faster rate than americium and after 4 days there will be a significant loss of mass.

7 0

3 years ago

Why can’t there be a temperature lower than absolute zero

Firdavs [7]

Atoms cannot physically get more compact (temperature lowering increases the space between atoms) At absolute zero, the atoms are as physically close together as allowed by the physical laws governing it.

7 0

3 years ago

A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

$V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}$

$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}$

<em>B) Radius</em>

$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}$

$V = \frac{ 4}{3 }\pi r^{3}$

$r^{3} = \frac{3V }{4 \pi }\$

$r= \sqrt [3]{ \frac{3V }{4 \pi } }$

$r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}$

3 0

3 years ago

How many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?

lisov135 [29]

Moles of glucose = Molarity x volume solution
= 4.5 x 1.5
= 6.75 moles.

Hope this helps, have a great day ahead!

3 0

3 years ago

What is the formula for hydrogen iodide trihydrate

Natali [406]

HI.3H2O. This is the answer

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3 years ago