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3 years ago

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Why does a 4 carbon linear alkane, butane c4h10, have two more hydrogens ?

1 answer:

Scilla [17]3 years ago

3 0

Alkanes are hydrocarbons that only contain single bonds in them. A carbon can bond with up to 4 atoms, even with another carbon atom. So, in a C-C bond, 3 more H atoms can bond to each of the C atom. Generally, the chemical formula for alkanes is CₓH₂ₓ₊₂. So for butane, there are 4 C atoms. The corresponding H atoms are 2(4) + 2 = 10. That's why it's chemical formula is C₄H₁₀.

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The moon lost its volcanic __ beacuse of its weak gravity

matrenka [14]

Answer:

Explanation:

the moon is still volcanic

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3 years ago

Digiron [165]

First we need to find the number of moles that 43.9g of gallium metal is. We can do this by finding the molar weight of gallium and cross-multiplying to cancel out units:

$\frac{1mole}{69.723g} *\frac{43.9g}{1} =0.63MOLga$

So we are dealing with 0.63 moles of gallium metal.

We can take from the balanced equation that 4 moles of gallium metal will react completely with 3 moles of oxygen gas. We can take this ratio and make a proportion to find the amount of oxygen gas, in moles, that will react completely with 0.63 moles of gallium metal:

$\frac{4molesGa}{3moles02} =\frac{0.63molesga}{xmoleso2}$

Cross multiply and solve for x:

4x=1.89

x=47 molesO₂

So now we know that 0.47 moles of oxygen gas will react with 43.9g of gallium metal.

7 0

3 years ago

What reactants go into the process?<br> I

Eva8 [605]

Answer:

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Explanation:

7 0

3 years ago

Read 2 more answers

Ill upvote please help

NARA [144]

Hello there.

<span>Find the average acceleration of a northbound subway train that slows from 12 m/s to 9.6 m/s in 0.9 s

</span><span>2.7 m/s2
</span>

3 0

4 years ago

A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin

lorasvet [3.4K]

Answer : The mass of the water in two significant figures is, $3.0\times 10^1g$

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron metal = $0.45J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of iron metal = 32.3 g

$m_2$ = mass of water = ?

$T_f$ = final temperature of mixture = $59.2^oC$

$T_1$ = initial temperature of iron metal = $21.9^oC$

$T_2$ = initial temperature of water = $63.5^oC$

Now put all the given values in the above formula, we get

$32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC$

$m_2=30.16g\approx 3.0\times 10^1g$

Therefore, the mass of the water in two significant figures is, $3.0\times 10^1g$

3 0

3 years ago