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4 years ago

Which kind of nuclear decay does not alter the identity of the atom?

Chemistry

2 answers:

Aloiza [94]4 years ago

8 0

The answer would be
Gamma Radiation

alukav5142 [94]4 years ago

4 0

Hmmmmmm, Gamma Radiation

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Calculate the energy required to heat 566.0mg of graphite from 5.2°C to 23.2°C. Assume the specific heat capacity of graphite un

maw [93]

Answer:

7.23 J

Explanation:

Step 1: Given data

Mass of graphite (m): 566.0 mg

Initial temperature: 5.2 °C

Final temperature: 23.2 °C

Specific heat capacity of graphite (c): 0.710J·g⁻¹K⁻¹

Step 2: Calculate the energy required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.710J·g⁻¹K⁻¹ × 0.5660 g × (23.2°C-5.2°C)

Q = 7.23 J

6 0

3 years ago

How does the atmosphere help to make Earth livable? Check all that apply.

Likurg_2 [28]

Answer:

- contains oxygen

- provides warmth

- blocks harmful energy from the Sun

Explanation:

The atmosphere is crucial when it comes to providing living conditions on Earth. The gases in the atmosphere manage to balance the things just about right, so that the life can thrive. The atmosphere has oxygen, the gas that is used by pretty much every organism to be able to function. With its greenhouse gases, the atmosphere manages to trap the heat inside, thus providing warmth to the Earth. The numerous layers of the atmosphere manage to block in big portion the harmful energy from the Sun, with a special accent on the ozone layer, a layer that manages to stop most of the UV radiation.

5 0

3 years ago

Read 2 more answers

How many values of these of these trials can be considered accurate?

Gnoma [55]

Answer:

go to google

Explanation:it all ways work

6 0

3 years ago

Convert the Haworth projection of the carbohydrate below into its corresponding Fischer projection (in the standard format with

Harlamova29_29 [7]

Answer:

See explaination

Explanation:

H O / I MOH the OH da Anomer My nequatorial bovel SOH niematerial and CH2OH Yr axial boud - H o H o . H (Auro) 2 flip . note: 1)Fischer projection : L- Sugar ( OH group is in the left side ) at C-5 carbon (bottom most chiral center )

In the Haworth projection given above in the qsn it is Beta anomer( -CH2OH & C-OH are in same side i.e. Cis to each other ). It is a L -sugar .

I omitted C2 -H bond (axial ) in the chair conformation of alpha anomer.

See attachment for further solution

8 0

3 years ago

Using the table below, identify the unknown material that has a mass of 15,06 g and a

kati45 [8]

The unknown material : gold

<h3>Further explanation </h3>

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

ρ = density

m = mass

v = volume

mass of unknown the material : 15.06 g

volume = 0.78 ml

The density :

$\tt \rho=\dfrac{m}{V}\\\\\rho=\dfrac{15.06}{0.78}=19.31~g/ml$

materials that match the density: gold

8 0

3 years ago