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tekilochka [14]
1 year ago
10

HELP!! ASAP!! PLEASE!!

Chemistry
1 answer:
kvv77 [185]1 year ago
6 0

The bonds that are present in the pictured compound are C −C and C −H.

<h3>What is a bond?</h3>

A bond is that which connects two atoms together. We know that a compound is composed of bonds. There are a number of bonds in a compound. The bonds are always as many as the atoms that need to be linked in the compound.

We know that butane a shown in the image is an alkane. An alkane has carbon - carbon and hydrogen - carbon bonds. There are no multiple bonds in the compound butane.

Thus, the bonds that are present in the pictured compound are C −C and C −H.

Learn more about chemical bond:brainly.com/question/15444131

#SPJ1

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What would happen if you play a note in a cold day on a flute?
Maurinko [17]
I don't know if this is the answer you are looking for but it would be flat unless the player pushed the tuning slide in.
7 0
3 years ago
Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o
Sunny_sXe [5.5K]

Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}

Now, we are able to solve for m:

m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} }  = 9.70*10^4 g Na=97 Kg Na=97 000 g Na

7 0
3 years ago
Read 2 more answers
An aqueous solution containing 17.5 g of an unknown molecular (nonelectrolyte) compound in 100.0 g of water has a freezing point
Sonbull [250]

Answer:

molar mass = 180.833 g/mol

Explanation:

  • mass sln = mass solute + mass solvent

∴  solute: unknown molecular (nonelectrolyte)

∴ solvent: water

∴ mass solute = 17.5 g

∴ mass solvent =  100.0 g = 0.1 Kg

⇒ mass sln = 117.5 g

freezing point:

  • ΔTc = - Kc×m

∴ ΔTc = -1.8 °C

∴ Kc H2O = 1.86 °C.Kg/mol

∴ m: molality (mol solute/Kg solvent)

⇒ m = ( - 1.8 °C)/( - 1.86 °C.Kg/mol)

⇒ m = 0.9677 mol solute/Kg solvent

  • molar mass (Mw) [=] g/mol

∴ mol solute = ( m )×(Kg solvent)

⇒ mol solute = ( 0.9677 mol/Kg) × ( 0.100 Kg H2O )

⇒ mol solute = 0.09677 mol

⇒ Mw solute = ( 17.5 g ) / ( 0.09677 mol )

⇒ Mw solute = 180.833 g/mol

6 0
2 years ago
The following are formed as white precipitates in solution except
aniked [119]

Answer:

A. Iron (III) hydroxide

Explanation:

Orange-brown or rust

6 0
2 years ago
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A workman uses a moveable pulley to lift a heavy load of bricks. What is the input force, if the output force is 150 N?
NeTakaya

Answer:

the input force would be 75 N

Explanation:

if the output force is 150 N you divide that in half which leaves with 75 N !! :)

7 0
3 years ago
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