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3 years ago

What is the answers?

Chemistry

1 answer:

ra1l [238]3 years ago

5 0

the numbers are going to be small so like a power but its at the bottom

NH3, H2O2, NHO2

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Which of the following factors will affect the freezing point of a solution?

erastovalidia [21]

The freezing point of the solvent in a solution changes as the concentration of the solute in the solution changes (but it does not depend on the identity of either the solvent or the solute(s) particles (kind, size or charge) in the solution).

Generally, pressures lower than 1 atmosphere lower the temperature at which a substance freezes, but for water, a higher pressure gives a lower freezing point. The force from a pressure change figures into the molecular forces already at play in a substance.

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3 years ago

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3 years ago

Elements x and y form 2 binary compounds. In the first 14.0 g of x combines with 3.0 g of y. In the second, 7.00 g of x combine

Roman55 [17]

second compound

Let molar mass of x is = X

Let molar mass of y is = Y

Moles of x in second compound = Mass / molar mass = 7 / X

Moles of y in second compound = Mass / molar mass = 4.5 / Y

For second compound

7 / X : 4.5/ Y = 1:1

Therefore

X / Y = 7/4.5

Y / X = 4.5/ 7

The mass of x in first compound = 14g

moles of x in first compound = 14/X

Mass of y in first compound = 3

moles of y in first compound = 3 / Y

14 / X : 3/ Y = 14Y / 3X = 14 X 4.5 / 3 X 7 = 3 :1

Thus molar ratio in first compound = moles of x / Moles of y = 3:2

Formula = x3y

6 0

3 years ago

Que significa arte m m k,

Svetradugi [14.3K]

Answer:

Explanation:

huh

7 0

3 years ago

According to the following reaction, how many grams of potassium phosphate will be formed upon the complete reaction of 29.6 gra

Ugo [173]

Answer:

There is 37.36 grams of K3PO4 produced

Explanation:

Step 1: Data given

Mass of H3PO4 = 29.6 grams

KOH is in excess

Molar mass of KOH = 56.11 g/mol

Molar mass of H3PO4 = 97.99 g/mol

Step 2: The balanced equation

3KOH(aq) + H3PO4(aq) ⇔ K3PO4(aq)+3H2O(l)

Step 3: Calculate mass of KOH

Mass KOH = mass KOH / molar mass KOH

Mass KOH = 29.6 grams / 56.11 g/mol

Mass KOH = 0.528 moles

Step 4: Calculate moles of K3PO4

Since KOH is the limiting reactant, We need 3 moles of KOH for each moles of H3PO4, to produce 1 mole of K3PO4 and 3 moles of H2O

For 0.528 moles of KOH we'll have 0.528/3 = 0.176 moles of K3PO4

Step 5: Calculate mass of K3PO4

Mass K3PO4 = moles K3PO4 * molar mass K3PO4

Mass K3PO4 = 0.176 moles * 212.27 g/mol

Mass K3PO4 = 37.36 grams

There is 37.36 grams of K3PO4 produced

8 0

3 years ago