A squirrel weighs 3,750 grams, how much does the squirrel weigh in hectograms?

Which of the following are gamma rays NOT used for?`A. None of theseB. Industrial imagingC. Medical imagingD. Sterilization

kompoz [17]

Answer

A. None of these

Explanation

Gamma rays are ionizing electromagnetic radiation, obtained by the decay of an atomic nucleus. Gamma rays are more penetrating, in the matter, and can damage living cells to a great extent. Gamma rays are used in medicine (radiotherapy), industry (sterilization and disinfection), and the nuclear industry.

Thus, gamma rays are used for all the options given in B - D. So the correct answer is option A. None of these

3 0

1 year ago

A chemist is looking at an electrically neutral atom that has three valence electrons. If she knows the element is a metalloid,

Rus_ich [418]

Your answer is boron,

I looked at the periodic table, there is nothing else that falls under this specs.

8 0

3 years ago

10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C

Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

$Q=mc\Delte T$

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c = 0.385 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC$

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c = 0.903 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC$

3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c = 2.42 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC$

4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c = 4.18J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC$

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c = 0.128 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC$

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

3 0

4 years ago

Chemistry help will be grateful thanks

igor_vitrenko [27]

Answer:

a = double displacement(iii)

b = Single displacement (ii)

c = Synthesis reaction (i)

d = decomposition reaction (iv)

Explanation:

It's not that hard!

6 0

3 years ago

Round off each measurement to the number of significant figures shown in parentheses.

Wewaii [24]

We have to round off each measurement to the number of significant figures.

The correct answers are: 265.9, 0.00045 and 65 respectively.

265.991 when expressed in 4 significant figure it rounds off as 265.9 because 265.9 represents the value 265.991.

0.0004513 when expressed in 2 significant figure it rounds off as 0.00045 because zero before and after decimal are not significant.

6593 when expressed in 2 significant figure it rounds off as 65.

6 0

3 years ago