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kotegsom [21]
2 years ago
12

An experiment requires that each student use an 8.5 cm length of magnesium ribbon. How many students can do the experiment if th

ere is a 570 cm length of magnesium ribbon available?
Chemistry
2 answers:
melisa1 [442]2 years ago
7 0
570/8.5=67.0 58... you only have to take the natural part, si the answer is 67 students
BlackZzzverrR [31]2 years ago
5 0

In the particular experiment, each student uses 8.5 cm of ribbon.

The total length of the Mg ribbon present is 570 cm

so to find how many students can use the ribbon

number of students = total length of the ribbon / length each student uses

number of students = 570 cm / 8.5 cm/student = 67. 1 students

since it should be a whole number it should be 67

therefore 67 students can use the ribbon

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Industrial production of nitric acid, which is used in many products including fertilizers and explosives, approaches 10 billion
mylen [45]

Answer: 9.361\times 10^{4} kJ

Explanation:

The balanced chemical equation :

4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)  \Delta H^0_{rxn}=-902.0kJ

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{7.056\times 10^3g}{17g/mol}=415.1moles

According to stoichiometry:

4 moles of NH_3 produces = 902.0 kJ of energy

415.1 moles of NH_3 produces =\frac{902.0}{4}\times 415.1=9.361\times 10^{4} kJ of energy

Thus the change in enthalpy is 9.361\times 10^{4} kJ

5 0
2 years ago
The atmosphere supports a column of mercury that is 748 mm in height. What is atmospheric pressure in torr? Convert this pressur
gulaghasi [49]

Answer:

748 torr

Explanation:

mmHg and torr are equivalent so, you'll have 748 torr.

7 0
2 years ago
The volume of a gas at 7.00°c is 49.0 ml. if the volume increases to 74.0 ml and the pressure is constant, what will the tempera
marshall27 [118]
Using charles law
v1/t1=v2/t2
v1=49ml
v2=74
t1=7+273=280k
t2=?
49/280=74/t2
0.175=74/t2  cross multiply
0.175t2=74
t2=74/0.175
t2=422k or 149celcius
8 0
2 years ago
40pionts
Pachacha [2.7K]

Problem One (left)

This is just a straight mc deltaT question

<em><u>Givens</u></em>

m = 535 grams

c = 0.486 J/gm

tf = 50

ti = 1230

Formula

E = m * c * (ti - tf)

Solution

E = 535 * 0.486 * ( 1230 - 50)

E = 535 * 0.486 * (1180)

E = 301077

Answer: A

Problem Two

This one just requires that you multiply the two numbers together and cut it down to 3 sig digits.

E = H m

H = 2257 J/gram

m = 11.2 grams

E = 2257 * 11.2

E = 25278  to three digits is 25300 Joules. Anyway it is the last one.

Three

D and E are both incorrect for the same reason. The sun and stars don't contain an awful lot of Uranium (1 part of a trillion hydrogen atoms). It's too rare. The other answers can all be eliminated because U 235 is pretty stable in its natural state. It has a high activation complex.

Your best chance would be enriched Uranium (which is another way of saying refined uranium). That would be the right environment. Atomic weapons and nuclear power plants (most) used enriched Uranium. You can google "Little Boy" if you want to know more.

Answer: B

Four

The best way to think about this question is just to get the answer. Answer C.

A: incorrect. Anything sticking together implies a larger and larger result. Gases don't work that way. They move about randomly.

B: Wrong. Heat and Temperature especially depend on movement. Stopping is not permitted. If a substance's molecules stopped, the substance would experience an extremely uncomfortable temperature drop.

C: is correct because the molecules neither stop nor do they stick. The hit and move on.

D: Wrong. An ax splitting something? That is not what happens normally and not with ordinary gases. It takes more energy that mere collisions or normal temperatures would provide to get a gas to split apart.

E: Wrong. Same sort of comment as D. Splitting is not the way these things work. They bounce away as in C.

Five

Half life number 1 would leave 0.5 grams behind.

Half life number 2 would leave 1/2 of 1/2 or 1/4 of the number of grams left.

Answer: 0.25

Answer C

6 0
2 years ago
A student is given two 10g samples, each a mixture of only NaCl(s) and KCl(s) but in different proportions. Which of the followi
8090 [49]

Answer:

The information that can be used to determine which mixture has the higher proportion of KCl IS INFORMATION ABOUT THE MASS OF CHLORINE IN EACH MIXTURE, THIS INFORMATION CAN BE OBTAINED BY USING THE LAW OF DEFINITE PROPORTION.

Explanation:

The law of definite proportion states that the chemical composition by mass of a chemical compound is always constant. For instance, a chemical compound that is made up of two elements will always contain the same proportions of the constituent elements regardless of the quantity of chemical that was used.

Using the law of definite proportion, we can determine the proportion of sodium and chlorine in NaCl and the proportion of potassium and chlorine in KCl if the mass of chlorine that was used is known. Based on the results obtained, one can easily determine the mixtures that has higher proportion of KCl.

6 0
2 years ago
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