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3 years ago

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Which is true of ultraviolet rays?

1 answer:

Ket [755]3 years ago

7 0

A is true of UV rays.
B is true not of UV rays but rather of visible light.
C is true not of UV rays but rather of microwaves. (unless you actually toast your toast in a toaster like a normal person)
D is true not of UV rays but rather of radio waves.

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Ima 5% glucose solution, how many grams of glucose would be present per 100mL

Sedaia [141]

Percentage Weight-in-volume is defined as the <em><u>number of grams of a solute in a 100 ml (milliliters) solution.</u></em>

<u />

<u>Percentage Weight-in-volume</u> can tell us about the <em>degree of concentration of a given solution.</em>

<em><u /></em>

The solute can be <em>crystalline or non-crystalline in nature.</em>

<em></em>

The <u>number of grams of glucose</u> present in a <u>5% glucose solution</u> is 5 grams.

This question is based on a Percentage Weight-in-volume. The formula states that:

a% of a glucose solution =<u> a grams of glucose in a 100 mL solution</u>

Hence, 5% glucose solution = 5 grams of glucose / 100 mL solution

Therefore, the <u>number of grams of glucose</u> present in a <u>5% glucose solution</u> is 5 grams.

To learn more, visit the link below:

brainly.com/question/8482854

6 0

2 years ago

A 20.0 mL of 0.27 M solution of the salt NaA has a pH of 8.70.

zlopas [31]

Answer:

a) 3.969

b) 3.489

Explanation:

a) Calculate the pk, value of the acid HA

PH of salt hydrolysis

P $^{H}$ = 1/2 ( pkw + pka + logC )

8.7 * 2 = 14 + log ( 0.27 ) + Pka

∴ Pka = 3.9686 ≈ 3.969

b) Calculate the PH of a solution containing 0.3 M HA and 0.1 M NaA

PH = Pka + log ( salt / acid )

= 3.9686 + log ( 0.1 / 0.3 )

= 3.9686 - 0.48 = 3.489

3 0

3 years ago

The study of all chemicals containing carbon

laila [671]

The correct answer is
<span>3. Organic Chemistry.....</span>

6 0

3 years ago

Read 2 more answers

A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate for

laiz [17]

Answer:

CaF2 will not precipitate

Explanation:

Given

Volume of Ca(NO3)2 $= 10$ ml

Molar concentration of Ca(NO3)2 $= 0.001$

Volume of NaF $= 10$ ml

Molar concentration of NaF $= 0.0001$

Ksp for CaF2 $= 3.2 * 10^ {-11}$

CaF2 will precipitate if Q for the reaction is greater than ksp of CAF2

Moles of calcium ion

$= 10 * 0.001\\= 0.01$

$[Ca2+] = \frac{0.01}{10 + 10} \\= \frac{0.01}{20} \\= 5 * 10^{-4}$

Moles of F- ion

$= 10 * 0.0001\\= 0.001$

$[F-] = \frac{0.001}{10 + 10} \\= \frac{0.001}{20} \\= 5 * 10^{-5}$

$Q = [Ca2+] [F-]^2\\= (5 * 10^{-4}) * (0.5* 10^-4)\\= 1.25 * 10^{-12}$

Q is lesser than Ksp value of CaF2. Hence it will not precipitate

5 0

3 years ago

What is the density of heavy water?

vlada-n [284]

<span>1.11 g/cm³

Hope this helps! </span>

6 0

3 years ago