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Arisa [49]
3 years ago
8

Ethylene glycol has a density of 1.11g/cm3. What is the mass in grams of 387ml of ethylene glycol?

Chemistry
1 answer:
FromTheMoon [43]3 years ago
4 0
Density is defined as the mass per unit volume. 
formula is as follows
density =  \frac{mass}{volume}
we have been given the density and volume so we need to find the mass of ethylene glycol
if we rearrange the equation making mass the subject
mass = density x volume
density = 1.11 g/cm³
this means that 1 cm³ of solution has a mass of 1.11 g
since mL = cm³
therefore 387 cm³ of solution has a mass of 1.11 g/cm³ x 387 cm³ = 429.6 g
mass of ethylene glycol is 429.6 g
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7 0
3 years ago
PLSSSSSSSS help !!!!! One mole of an element contains which of the following?
irina [24]

Answer:

6.02 × 10²³ atoms

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For example:

32 g of oxygen = one mole = 6.02 × 10²³  atoms O.

1.008 g of hydrogen = one mole = 6.02 × 10²³  atoms of H.

or

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5 0
3 years ago
Find the number of grams <br><br>4.00 moles of CU(CN)2 ​
USPshnik [31]

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462g

Explanation:

First, let us calculate the molar mass of Cu(CN)2. This is illustrated below:

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4 0
2 years ago
When an aqueous solution of AgNO3 is electrolyzed, a gas is observed to form at the anode. The gas is
iris [78.8K]

Answer:

Oxygen

Explanation:

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In the anode, the substance lose electrons to undergo oxidation.

From the 4 ions, only OH- can lose electrons to form water and oxygen,

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While others tend to gain electrons to form new substances instead (they undergo reduction).

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6 0
3 years ago
Hc2h3o2(aq)+h2o(l)⇌h3o+(aq)+c2h3o−2(aq) kc=1.8×10−5 at 25∘c part a if a solution initially contains 0.260 m hc2h3o2, what is the
Cloud [144]

Answer : The equilibrium concentration of H_3O^+ at 25^oC is, 2.154\times 10^{-3}m.

Solution : Given,

Equilibrium constant, K_c=1.8\times 10^{-5}

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Let, the 'x' mol/L of H_3O^+ are formed and at same time 'x' mol/L of HC_2H_3O_2 are also formed.

The equilibrium reaction is,

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The expression for equilibrium constant for a given reaction is,

K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

Now put all the given values in this expression, we get

1.8\times 10^{-5}=\frac{(x)(x)}{(0.260-x)}

By rearranging the terms, we get the value of 'x'.

x=2.154\times 10^{-3}m

Therefore, the equilibrium concentration of H_3O^+ at 25^oC is, 2.154\times 10^{-3}m.

4 0
2 years ago
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