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3 years ago

What is the wavelength of the above wave?

Chemistry

1 answer:

svet-max [94.6K]3 years ago

5 0

Answer:

well what above wave?? do you have a picture

Explanation:

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SOS!!!You are trying to determine the volume of the balloon needed to match the density of the air in the lab. You know that if

guapka [62]

The density of air surrounding the balloon may be calculated by assuming that air is an ideal gas and using the equation,
PV = nRT
In terms of density,
density = PM/RT
Using the given data,
density = (1 atm)(28.84 g/mol)/(0.0821 L.atm/molK)(20+273K) = 1.20 g/L

Using ratio and proportion,
density of balloon = 0.577 g/ V = 1.20 g/L

The volume is equal to 0.48 L.

7 0

4 years ago

How many cobalt atoms are in 345 g of cobalt

lilavasa [31]

Answer:

$\boxed {\boxed {\sf About \ 3.53 \ *10^{24}atoms \ Co}}$

Explanation:

To convert from grams to atoms:

Convert grams to moles
Convert moles to grams

1. Convert grams to moles

First, find the molar mass of cobalt using the Periodic Table of Elements.

Cobalt (Co): 58.93319 g/mol

Next, use this mass as a ratio or fraction.

$\frac{58.93319 \ g\ Co}{1 \ mol \ Co}$

Multiply the mass of the given sample (345 grams) by this ratio.

$345 \ g \ Co *\frac{58.93319 \ g\ Co}{1 \ mol \ Co}$

Flip the fraction so the grams of Cobalt will cancel each other out when multiplying.

$345 \ g \ Co *\frac{1 \ mol \ Co}{58.93319 \ g\ Co}$

$345 \ *\frac{1 \ mol \ Co}{58.93319 }$

$\frac{345 \ mol \ Co}{58.93319 } =5.854086636 \ mol \ Co$

2. Convert moles to atoms

Use Avogadro's Number, which tells us the number of units (in this case atoms) in 1 mole.

$6.022 *10^{23} \ atoms/mol$

Use this number as a ratio or fraction.

$\frac{6.022 * 10^{23} \ atoms \ Co}{1 \ mol \ Co}}$

Multiply this ratio by the number of moles we found.

$5.854086636 \ mol \ Co*\frac{6.022 * 10^{23 }\ atoms \ Co}{1 \ mol \ Co}}$

The moles of Cobalt will cancel.

$5.854086636 \ *\frac{6.022 * 10^{23 }\ atoms \ Co}{1 }}$

$5.854086636 \ *{6.022 * 10^{23 }\ atoms \ Co}$

$3.52533097*10^{24} \ atoms \ Co$

3.Round

The original measurement of 345 has 3 significant figures (3, 4, and 5). We must round to 3 sig figs, which is the hundredth place for this measurement.

$3.52533097*10^{24} \ atoms \ Co$

The 5 in the thousandth place tells us to round the 2 to a 3.

$3.53 \ *10^{24}atoms \ Co$

There are about 3.53*10²⁴ atoms of cobalt in 345 grams.

6 0

3 years ago

correctly complete the following statement: The side chains ('-R groups') of amino acids in a beta-sheet: A. alternate pointing

Arlecino [84]

Answer:

Explanation:

The β-sheet is a twisted pattern in which the protein strands are laterally linked through hydrogen bonds. These β-sheet motifs are segments of 3 to 10 amino acids that are specially configured to form beta antiparallel strands. In β-sheet regions, the hydrogen bonds are localized among carbonyl and amino groups of the polypeptide backbone, and side chains (i.e., the R groups) are extended up and down in the plane of the β-sheet.

6 0

4 years ago

As the depth of the ocean increases the amount of sunlight and water temperature will..

Harlamova29_29 [7]

Answer:

Decrease

Explanation:

The lower you go, the colder because the sun can't reach lower levels of the ocean.

4 0

3 years ago

Read 2 more answers

100.0 mL of a 0.780 M solution of KBr is diluted to 500.0 mL. What is the new concentration of the solution?

Ainat [17]

5 times dilution
0.780M x 1/5 = 0.156M
Hope this help.

8 0

3 years ago