Name the hydrocarbon...i need help

Chemistry

1 answer:

kondaur [170]2 years ago

4 0

Answer:

The given hydrocarbon is Ethylene Glycol

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***BRAINLIEST ASNWERRR***<br>How many grams are in 34.2 moles of Lithium (Li)?

Step2247 [10]

Mass of Li = 237.38 g

<h3>Further explanation</h3>

The mole itself is the number of particles contained in a substance amounting to 6.02.10²³

$\large {\boxed {\boxed {\bold {mol = \frac {mass} {molar \: mass}}}}$

<h3>Known</h3>

Moles of Li = 34.2

Molar mass(MW) of Li = 6.941 g/mol

then mass of Lithium (Li) :

$\tt mol=\dfrac{mass}{MW}\\\\mass=mol\times MW\\\\mass=34.2\times 6,941~g/mol\\\\mass=\boxed{\bold{237.38~g}}$

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2 years ago

Choose all that apply. Nitrogen has an atomic number of 7. Its atomic mass is 14. Nitrogen has

Goshia [24]

The atomic number of calculated as number of protons + number of neutrons. The number of protons are the same as atomic number. This implies that Nitrogen has seven protons. To get the the number of neutron we minus atomic mass from number of protons
that is 14-7=7 neutrons

6 0

2 years ago

How the proton H gradient is used to make ATP

sladkih [1.3K]

Answer:

The proton gradient produced by proton pumping during the electron transport chain is used to synthesize ATP. Protons flow down their concentration gradient into the matrix through the membrane protein ATP synthase, causing it to spin (like a water wheel) and catalyze conversion of ADP to ATP.

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2 years ago

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

irinina [24]

Answer:

$m_{KBr}=6.030gKBr$

Explanation:

Hello.

In this case, since the normal boiling point of X is 117.80 °C, the boiling point elevation constant is 1.48 °C*kg*mol⁻¹, the mass of X is 100 g and the boiling point of the mixture of X and KBr boils at 119.3 °C, we can use the following formula:

$(T_b-T_b_0)=i*m*K_b$

Whereas the Van't Hoff factor of KBr is 2 as it dissociates into potassium cations and bromide ions; it means that we can compute the molality of the solution:

$m=\frac{T_b-T_b_0}{i*K_b}=\frac{(119.3-117.8)\°C}{2*1.48\°C*kg*mol^{-1}}\\ \\m=0.507mol/kg$