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ch4aika [34]
3 years ago
6

If there are no chemical controls which can be implemented at your worksite, what can you do to protect yourself?

Chemistry
1 answer:
saul85 [17]3 years ago
5 0
the answer is D.) wear recommended PPE for protection
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When a rock falls from outer space all the way to the groung, its called a?
goblinko [34]
It's called a meteorite.
6 0
3 years ago
What does AS> O mean?
Vlad1618 [11]

Answer:

ΔS> 0 means Letter A

Explanation:

Processes that involve an increase in entropy of the system (ΔS > 0) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include the surroundings, we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true:

\displaystyle \Delta {S}_{\text{univ}}=\Delta {S}_{\text{sys}}+\Delta {S}_{\text{surr}}

5 0
3 years ago
Which set of coefficients will balance this chemical equation?
Lena [83]

Answer:The answer is D.1,3

Explanation:

8 0
3 years ago
How many liters of 1.75 M solution could be made using 35 grams of NaCl?
dolphi86 [110]
Data:
M (molarity) = 1.75 M (mol/L)
m (mass) = 35 g
MM (molar Mass) of NaCl = 58.44 g/mol
V (volume) = ? (in liters)

Formula:
M =  \frac{m}{MM*V}

Solving:
M = \frac{m}{MM*V}
1.75 =  \frac{35}{58.44*V}
1.75*58.44V = 35
102.27V = 35
V =  \frac{35}{102.27}
\boxed{\boxed{V \approx 0.34\:L}}\end{array}}\qquad\quad\checkmark
3 0
3 years ago
When the pressure that a gas exerts on a sealed container changes from 1100 bar to 75.5 bar, the temperature changes from k to 2
Serjik [45]
Gay-Lussac's law gives the relationship between pressure and temperature of gas. For a fixed amount of gas, pressure is directly proportional to temperature at constant volume.
P/T = k
where P - pressure , T - temperature and k - constant
\frac{P1}{T1} =  \frac{P2}{T2}
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation 
substituting the values in the equation 
\frac{1100 bar}{T}  =  \frac{75.5 bar}{298 K}
T = 4342 K
initial temperature was 4342 K
7 0
3 years ago
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