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kodGreya [7K]
3 years ago
14

Why might the ability to produce three-dimensional images of organs be helpful in diagnosing a patient's ailment?

Chemistry
1 answer:
seraphim [82]3 years ago
4 0

Answer:

Explanation: The body is largely composed of water molecules. Each water molecule has two hydrogen nuclei or protons. MRI takes advantage of the high prevalence of hydrogen in the body and the magnetic properties of the proton in a hydrogen atom. Hydrogen atoms induce a small magnetic field due to the spin of this atom's proton.

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A _______ pattern organizes a speech by incorporating repetition and variations of themes and ideas
astraxan [27]

Answer:

A wave pattern organizes a speech

8 0
3 years ago
What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP
Dvinal [7]
Vol.250 before its to much pressure

7 0
3 years ago
Read 2 more answers
If the titrant has a molarity of 0.1000 m and there are 45.00 ml of analyte present, what is the molarity of the analyte?
marysya [2.9K]
Given: 

Concentration of titrant = 0.1000 M
Volume of titrant = 45 mL

The molarity of analyte depends on the amount of the analyte present in the titrated solution. If the amount of analyte is 20 mL, then its concentration is:

45ml * 0.10 M = C analyte * 20 ml

C analyte = 0.225 M
4 0
2 years ago
24. What volume of a 0.0200M calcium hydroxide is required to neutralize 35.00 mL of 0.0500M nitric acid
Natalka [10]

Answer:

THE VOLUME OF 0.200M CALCIUM HYDROXIDE NEEDED TO NEUTRALIZE 35 mL of 0.050 M NITRIC ACID IS 43.75 mL.

Explanation:

Using

Ca VA / Cb Vb = Na / Nb

Ca = 0.0500 M

Va = 35 mL

Cb = 0.0200 M

Vb = unknown

Na = 2

Nb = 1

Equation for the reaction:

Ca(OH)2 + 2HNO3 --------> Ca(NO3)2 + 2H2O

So therefore, we make Vb the subject of the equation and solve for it

Vb = Ca Va Nb / Cb Na

Vb = 0.0500 * 35 * 1 / 0.0200 * 2

Vb = 1.75 / 0.04

Vb = 43.75 mL

The volume of 0.02M calcium hydroxide required to neutralize 35 mL of 0.05 M nitric acid is 43.75 mL

6 0
3 years ago
What is the density of CHCL3 vapor at 1.00atm and 298K?
Advocard [28]

Answer:

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.

Explanation:

By ideal gas equation:

PV=nRT

Number of moles (n)

can be written as: n=\frac{m}{M}

where, m = given mass

M = molar mass

PV=\frac{m}{M}RT\\\\PM=\frac{m}{V}RT

where,

\frac{m}{V}=d which is known as density of the gas

The relation becomes:

PM=dRT    .....(1)

We are given:

M = molar mass of chloroform= 119.5 g/mol

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the gas = 298K

P = pressure of the gas = 1.00 atm

Putting values in equation 1, we get:

1.00atm\times 119.5g/mol=d\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\d=4.88g/L

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.

7 0
3 years ago
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