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____ [38]
3 years ago
10

You should know the location of safety equipment

Chemistry
1 answer:
s2008m [1.1K]3 years ago
7 0
I would say you should use or test it once a week to ensure it is working properly in an active laboratory since it is a workplace with significant chemical hazards so it would give peace of mind to know on a quite regular basis that it can be relied on in case of an emergency like an eye flush for example.
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What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm o
lyudmila [28]

Answer:

\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu

Thus, the atom percent turns out:

\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%

Best regards.

4 0
3 years ago
CH4+O2—-CO2+H2O what is the best classification for the unbalanced equations reaction and why
yawa3891 [41]
I don’t know what you mean by classification exactly but it is a redox equation. The reactant side of carbon is losing hydrogen to form carbon dioxide. And oxygen is gaining hydrogen which gives you the water. Redox reactions are also known as combustion reactions.
5 0
3 years ago
Planetesimals are made from <br><br> A. Rock<br><br> B. Gas<br><br> C. Stars<br><br> D. Clouds
White raven [17]
A )rock is correct answer
3 0
3 years ago
Read 2 more answers
The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to produc
Elis [28]

Answer: 9.9 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{8.150g}{2g/mol}=4.08moles

b) moles of C_2H_4

\text{Number of moles}=\frac{9.330g}{28g/mol}=0.33moles

H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)

According to stoichiometry :

1 mole of C_2H_4 combine with 1 mole of H_2

Thus 0.33 mole of C_2H_4 will combine with =\frac{1}{1}\times 0.33=0.33 mole of H_2

Thus C_2H_4 is the limiting reagent as it limits the formation of product.

As 1 mole of C_2H_4 give =  1 mole of C_2H_6

Thus 0.33 moles of C_2H_4 give =\frac{1}{1}\times 0.33=0.33moles  of C_2H_6

Mass of C_2H_6=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9g

Thus theoretical yield (g) of C_2H_6 produced by the reaction is 9.9 grams

7 0
3 years ago
Which two terms relate and why? Compound, ion, proton
Lerok [7]

ion, proton

Explanation:

Ions and protons have profound relationships.

An ion is an atom that has lost or gained electrons.

A proton is a positively charged subatomic particle.

What is the relationship between an ion and a proton?

  • In an atom, there are three fundamental particles.
  • Protons are the positively charged particles located in the nucleus of atoms.
  • Electrons are the negatively charged particles orbiting round an atom.
  • Neutrons have no charges and they occupy the nucleus with protons.

Atoms are electrically neutral and this implies that they have equal number of protons and electrons.

In an ion, the number of protons and electrons differ.

For positive charged ions, the number of protons is more suggesting they have lost electrons.

For negatively charged ions. the number of protons is less suggesting they must have added electrons.

Compound are combinations of different atoms.

learn more:

Anions brainly.com/question/4670413

#learnwithBrainly

5 0
3 years ago
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