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soldi70 [24.7K]
3 years ago
11

Your friend cuts his sandwiches in half diagonally because then he gets twice as much

Chemistry
2 answers:
Tcecarenko [31]3 years ago
6 0

Answer:

It is still one sandwich it doesn't become twice the size just because you cut it, it just becomes one sandwich cut into two

Explanation:

daser333 [38]3 years ago
4 0
Wym? I dont have friends lol
You might be interested in
Write the Henderson-Hasselbalch equation for a propanoic acid solution ( CH3CH2CO2H , pKa=4.874 ) using the symbols HA and A− ,
zysi [14]

Answer:

a) [A⁻]/[HA] = 0.227

b) [A⁻]/[HA] = 0.991

c) [A⁻]/[HA] = 2.667

Explanation:

In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:

  • CH₃CH₂CO₂H = HA
  • CH₃CH₂CO₂⁻ = A⁻

pH = pka + Log [A⁻]/[HA]

pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]

  • (a)

4.23 = 4.874 + Log [A⁻]/[HA]

-0.644 = Log [A⁻]/[HA]

10^{-0.644} = [A⁻]/[HA]

0.227 = [A⁻]/[HA]

  • (b)

4.87 = 4.874 + Log [A⁻]/[HA]

-0.004 = Log [A⁻]/[HA]

10^{-0.004} = [A⁻]/[HA]

0.991 = [A⁻]/[HA]

  • (c)

5.30 = 4.874 + Log [A⁻]/[HA]

0.426 = Log [A⁻]/[HA]

10^{0.426} = [A⁻]/[HA]

2.667 = [A⁻]/[HA]

6 0
3 years ago
Which statement best describes the oxidation numbers of the atoms found in magnesium chloride? A. Magnesium has a 2- oxidation n
In-s [12.5K]
B. is the correct answer of your question
6 0
3 years ago
Read 2 more answers
Match the prefixes. 1. 1 hex- 2. 2 eth- 3. 3 prop- 4. 4 hept- 5. 5 non- 6. 6 dec- 7. 7 pent- 8. 8 but- 9. 9 meth- 10. 10 oct-
erma4kov [3.2K]

Hi!

All these prefixes are commonly employed in the nomenclature in organic chemistry. These are associated with organic compounds, such as alkanes, alkenes, alcohols and carboxylic acids -and these prefixes are known as word root.

<h3>The answers would be:</h3><h3>a) 1 - meth</h3>

Meth is associated with the number one, in that it signifies the presence of one carbon in the compound in concern. For instance, a one carbon alkane would be known as methane -with <em>ane </em>being indicative of the compound being an alkane, and meth denoting the presence of one carbon

<h3>b) 2 - eth</h3>

Eth is the associated prefix with the number 2 as it denotes the presence of 2 carbon atoms in the compound. For instance, the compound ethene gets it name for having 2 carbon atoms (denoted by eth). The eth acts as a prefix to the characteristic <em>ene</em> of alkenes.

<h3>c) 3 - prop</h3>

Prop is the prefix that signifies the presence of 3 carbons in the compound in concern. For instance, propanol is a three carbon alcohol, which we can tell by the name. Prop indicating three carbons in the compound, and <em>anol</em> being indicative of the compound being an alcohol.

<h3>d) 4 - but</h3>

But is associated with the number one, in that it denotes the presence of 4 carbons in the compound. For instance, a 4 carbon carboxylic acid would be known as butanoic acid -with anoic<em> </em>being indicative of the compound being a carboxylic acid, and but signifying the presence of four carbons.

<h3>e) 5 - pent</h3>

Pent is the associated prefix with the number five as it denotes the presence of five carbon atoms in the compound. For instance, the compound pentene gets it name for having five carbon atoms (denoted by pent). The pent acts as a prefix to the characteristic <em>ene</em> of alkenes.

<h3>f) 6 - hex</h3>

Hex is the prefix that signifies the presence of six carbons in the compound in question. For instance, a six carbon alkane would be known as hexane -with <em>ane </em>being indicative of the compound being an alkane, and hex denoting the presence of one carbon

<h3>g) 7 - hept</h3>

Hept is the associated prefix with the number seven as it denotes the presence of seven carbon atoms in the compound. For instance, the compound heptene gets it name for having seven carbon atoms (denoted by hept). The hept acts as a prefix to the characteristic <em>ene</em> of alkenes.

<h3>h) 8 - oct</h3>

Oct is the prefix that is associated with the number 8. This is because it denotes the presence of eight carbon atoms in the compound. For instance, the compound octane gets it name for having eight carbon atoms (denoted by oct). The oct acts as a prefix to the characteristic <em>ane</em> of alkanes.

<h3>i) 9 - non</h3>

Non is the prefix that is associated with the number nine, and thus denotes the presence of nine carbon atoms in the compound. For instance, for alkanes, a nonane would be a nine carbon atom alkane, with non denoting the presence of nine carbon atoms, and <em>ane</em> being indicative that the compound it an alkane.

<h3>j) 10- dec</h3>

Dec is the prefix that is associated with the number nine, and thus denotes the presence of nine carbon atoms in the compound. For instance, for alkanes, a decane would be an alkane with ten carbon atoms, with dec denoting the presence of ten carbon atoms, and <em>ane</em> being indicative that the compound it an alkane.

<h3>Hope this helps!</h3>
4 0
3 years ago
If the detector is capturing 3.3×108 photons per second at this wavelength, what is the total energy of the photons detected in
Reil [10]

Answer:

The total energy of the photons detected in one hour is 7.04*10⁻¹¹ J

Explanation:

The energy carried by electromagnetic radiation is displaced by waves. This energy is not continuous, but is transmitted grouped into small "quanta" of energy called photons. The energy (E) carried by electromagnetic radiation can be measured in Joules (J). Frequency (ν or f) is the number of times a wave oscillates in one second and is measured in cycles / second or hertz (Hz). The frequency is directly proportional to the energy carried by a radiation, according to the equation: E = h.f, (where h is the Planck constant = 6.63 · 10⁻³⁴ J / s).

Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. it is expressed in units of length (m). In light and other electromagnetic waves that propagate at the speed of light (c), the frequency would be equal to the speed of light (≈ 3 × 10⁸ m / s) between the wavelength :

f=\frac{speed of light}{wavelength}

So:

E=\frac{h*speed of light}{wavelength}

In this case, the wavelength is 3.35mm=3.35*10⁻³m and the energy per photon is:

E=\frac{6.63*10^{-34}*3*10^{8}}{3.35*10^{-3} }

E=5.93*10⁻²³ \frac{J}{proton}

The detector is capturing  3.3*10⁸ photons per second. So, in 1 hour:

E=5.93*10^{-23} \frac{J}{proton} *3.3*10^{8} \frac{proton}{s} *\frac{60}{1} \frac{s}{minute} *\frac{60}{1} \frac{minute}{hr}

E=7.04*10⁻¹¹ \frac{J}{hr}

The total energy of the photons detected in one hour is 7.04*10⁻¹¹ J

3 0
3 years ago
What is the electron configuration of the element with 27 protons?
adelina 88 [10]
<span>The electronic configuration of cobalt is :1s2 2s2 2p6 3s2 3p6 3d7 4s2 </span>
7 0
3 years ago
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