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AlexFokin [52]
3 years ago
6

Can someone please help asap!

Chemistry
1 answer:
posledela3 years ago
7 0
Mitosis is a stage of the cell cycle
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Changing the number of protons in an atom makes<br><br> A. an ion<br> B. an isotope
lana [24]
A. an ion

The atom gains a net electrical charge if the number of protons and electrons are not equal which makes it an ion.
5 0
3 years ago
What do the properties of alloys most resemble?
Tanzania [10]
The properties of alloys most resemble metal. Copper, tin, fiberglass.
6 0
3 years ago
What is the percentage by volume of 30.0 mL ethanol (C2H6O) dissolved in 150.0 mL water (H2O)?
NeX [460]

Answer: 16.7 %

Explanation:

\frac{30.0}{150.0+30.0}=\boxed{16.7\%}

5 0
2 years ago
Which of the following statements about the ordinary IR spectroscopic regions are TRUE? 1. In general, the IR FUNDAMENTAL region
yaroslaw [1]

Answer: the statements in 1 and 2 are true of IR spectroscopic region.

1. In general, the IR FUNDAMENTAL region has a longer wavelength region than the region we call the ultraviolet (uv) region.

2. We can sense some of the frequencies of the FUNDAMENTAL region of the IR as heat

Explanation:

IR has energy value between 10^-5eV - 10^-2eVwhile

UV has energy value of 4eV - 300eV

IR has low photon energy and cannot alter atoms and molecules while UV has sufficient energy to iodize atoms therefore UV has a higher energy band.

Infrared light falls just outside the visible spectrum, beyond the edge of what we can see as red.

5 0
3 years ago
what is the specific heat of a substance if 300 j are required to raise the temperature of a 267-g sample by 12 degrees c
slava [35]

Answer : The specific heat of the substance is 0.0936 J/g °C

Explanation :

The amount of heat Q can be calculated using following formula.

Q = m \times C \times \bigtriangleup T

Where Q is the amount of heat required = 300 J

m is the mass of the substance = 267 g

ΔT is the change in temperature = 12°C

C is the specific heat of the substance.

We want to solve for C, so the equation for Q is modified as follows.

C = \frac{Q}{m \times \bigtriangleup T}

Let us plug in the values in above equation.

C = \frac{300J}{267g \times 12 C}

C = \frac{300J}{3204 g C}

C = 0.0936 J/g °C

The specific heat of the substance is 0.0936 J/g°C

3 0
3 years ago
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