Need help asap!!!

I am a little confused

Sonbull [250]

The answer is D. I did that and i got it right.

7 0

3 years ago

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What is the molar mass of Na2CO3? 60.0 g/mol 106.0 g/mol 118.0 g/mol 141.0 g/mol

Dmitry [639]

Answer is: the molar mass od sodium carbonate (Na₂CO₃) is 106.0 g/mol.

M(Na₂CO₃) = 2 · Ar(Na) + Ar(C) + 3 · Ar(O).

M(Na₂CO₃) = 2 · 23 + 12 + 3 · 16 · g/mol.

M(Na₂CO₃) = 46 + 12 + 48 · g/mol.

M(Na₂CO₃) = 106 g/mol; molar mass of sodium carbonate.

Ar is relative atomic mass (the ratio of the average mass of atoms of a chemical element to one unified atomic mass unit) of an element.

8 0

3 years ago

What is 31 degrees Celsius in Fahrenheit

vovangra [49]

87.8 , 31 Celsius=87.8(88) in Fahrenheit

7 0

3 years ago

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The equation for the reaction between copper and nitric acid is vCu + wHNO3--->xCu(NO3)2 + yNO + zH2O Balance the equation an

lana [24]

Answer:

The coefficients of the balanced equation are:

v= 1
w= 4
x= 1
y= 2
z=2

Explanation:

The Law of Conservation of Matter, also called the Law of Conservation of Mass or the Lomonósov-Lavoisier Law, postulates that "mass is neither created nor destroyed, it is only transformed". This means that the reagents interact with each other and form new products with different physical and chemical properties than the reagents, but the amount of matter or mass before and after a transformation (chemical reaction) is always the same. In other words, then the mass before the chemical reaction is equal to the mass after the reaction.

Then, you must balance the following chemical equation:

v Cu + w HNO₃ → x Cu(NO₃)₂ + y NO + z H₂O

For that, you must first look at the subscripts next to each atom to find the number of atoms in the equation. If the same atom appears in more than one molecule, you must add its amounts

Left side: 1 copper, 1 hydrogen, 1 nitrogen and 3 oxygen.
Right side: 1 copper, 2 hydrogen, 3 nitrogen and 8 oxygen.

The coefficients located in front of each molecule indicate the amount of each molecule for the reaction. This coefficient can be modified to balance the equation, just as you should never alter the subscripts.

By multiplying the coefficient mentioned by the subscript, you get the amount of each element present in the reaction.

Generally, hydrogen and oxygen balance in the end. So you balance nitrogen first, because copper is already balanced (there is the same amount on both sides of the reaction). If w = 2, then:

Left side: 1 copper, 2 hydrogen, 2 nitrogen and 6 oxygen.
Right side: 1 copper, 2 hydrogen, 3 nitrogen and 8 oxygen.

But you see then that the oxygen is unbalanced and you have less quantity in the reagents. Then w must be greater. Being w = 4:

Left side: 1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.
Right side: 1 copper, 2 hydrogen, 3 nitrogen and 8 oxygen.

Going back to the idea of balancing nitrogen, being y = 2:

Left side: 1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.
Right side: 1 copper, 2 hydrogen, 4 nitrogen and 9 oxygen.

Balancing the hydrogen, being z = 2:

Left side: 1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.
Right side: 1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.

Since you have the same amount of each element on each side of the reaction, the reaction is balanced. Then, the balanced equation is:

Cu + 4 HNO₃ → Cu(NO₃)₂ + 2 NO + 2 H₂O

Finally:

v= 1
w= 4
x= 1
y= 2
z=2

6 0

3 years ago

Draw the structures of the aldol addition and condensation products of 3‑methylbutanal

Leona [35]

Aldol condensation involves two aldehydes or two ketones or an aldehyde and a ketone. The product of the reaction is shown in the image attached.

<h3>What is Aldol condensation?</h3>

Aldol condensation is a reaction that involves two aldehydes or two ketones or an aldehyde and a ketone. This reaction occurs at the carbonyl group as one of the carbonyl compounds is in the enolate form.

The product of the reaction is shown in the image attached. This gives the mechanism for the Aldol condensation of the compound 3‑methylbutanal.

Learn more about aldol condensation: brainly.com/question/9415260

6 0

2 years ago