TSNF AP Chemistry

TSNF AP Chemistry

10th - 12th Grade

25 Qs

quiz-placeholder

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TSNF AP Chemistry

TSNF AP Chemistry

Assessment

Quiz

Chemistry

10th - 12th Grade

Hard

NGSS
HS-PS1-1, HS-PS1-2, HS-PS1-7

+2

Standards-aligned

Created by

Charles Martinez

FREE Resource

25 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Media Image

According to the following reaction, how many grams of ammonia will be formed upon the complete reactions of 15.8 grams of nitrogen gas with excess hydrogen gas?

9.60 grams ammonia

19.2 grams ammonia

38.4 grams ammonia

4.80 grams ammonia

Tags

NGSS.HS-PS1-7

2.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

A sample of a hydrate of Na2CO3 with a mass of 550 grams was heated until all the water was removed. The sample was then weighed and found to have a mass of 300 grams. What is the formula for the hydrate?

Na2CO3 ⋅ 4H2O

Na2CO3 ⋅ 10H2O

Na2CO3 ⋅ 3H2O

Na2CO3 ⋅ 5H2O

Tags

NGSS.HS-PS1-7

3.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Al2O3(l) → Al(s) + O2(g)


Approximately how many liters of oxygen gas will be evolved at STP, if 25.68 grams of alumina decomposes?

22.4 L

8.46 L

5.64 L

11.2 L

Tags

NGSS.HS-PS1-7

4.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Media Image

The mass spectrum represented above is most consistent with which of the following elements?

Nb

Zr

Y

Mo

Tags

NGSS.HS-PS1-1

5.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

An element consists of 94.93% of an isotope with mass 31.97 u, 0.76% of an isotope with mass 32.97 u, 4.29% of an isotope with mass 33.97 u, and 0.02% of an isotope with mass 35.97 u. Calculate the average atomic mass, and identify the element.

32.06,S

32.07,P

3206,S

3207,P

6.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

The element boron (B) has two naturally occurring isotopes, 10B and 11B, with an average atomic mass of 10.812 u. Boron is 80.2 % 11B, and the atomic mass of 11B is 11.009 u. Calculate the mass of 10B.

10.5 u

11.01 u

10.01 u

10.8 u

7.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Media Image

The molecule drawn above has the same empirical formula as another compound "X" with a molar mass of 265.1.


Determine the molecular formula of compound "X".

C2H3O

C4H6O2

C12H18O6

C8H12O4

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