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Vikki [24]
3 years ago
10

Need help with some cemestery ​

Chemistry
1 answer:
Sergio [31]3 years ago
6 0

Explanation:

16 option c

17 option a

18 option a b

19 option d

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2.50 L of a gas at standard temperature and pressure is compressed to 575 mL. What is the new pressure of the gas
vfiekz [6]

Answer:

4.35 atm

Explanation:

According to the information given;

  • Initial volume of the gas, V₁ is 2.50 L
  • Initial pressure of the gas is standard pressure P₁, normally 1 atm
  • New volume of the gas, V₂ is 575 mL

We are required to determine the new pressure of the gas, P₂ ;

To answer the question, we are going to use the Boyle's law, that relates pressure and volume at constant temperature.

According to Boyle's law;

P₁V₁ = P₂V₂

Therefore, to determine the new pressure, P₂, we rearrange the formula;

New pressure, P₂ = P₁V₁ ÷ V₂

Thus;

P₂ = ( 1 atm × 2.50 L) ÷ 0.575 L

    = 4.3478 atm

     = 4.35 atm

Therefore, the new pressure of the gas is 4.35 atm

6 0
3 years ago
Read 2 more answers
What is any single living thing that can carry out life processes on its own?
ad-work [718]
Any single thing that can carry out life processes on its own is a dolphin.
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6 0
4 years ago
Identify the types of information that are necessary to communicate with emergency responders. Select one or more: Any chemicals
oksano4ka [1.4K]

Answer:

How the incident happened

Any chemicals involved in an incident

Any other hazards present in the lab

Explanation:

Above are the types of information that are necessary to communicate with emergency responders. The emergency responders ask the first question that how the incident happened. After that they ask that is there any harmful chemicals are present in the laboratory or what types of chemicals present in the laboratory. These questions were asked by the emergency responders in order to give the patient a suitable treatment.

8 0
3 years ago
What are the elements in the "dip" of the periodic table called?
pogonyaev

Answer:

Transition metals are the elements

3 0
4 years ago
A science teacher has a supply of 20% saline solution and a supply of 50% saline solution. How much of each solution should the
lilavasa [31]

Answer:

<h2>B)</h2>

Explanation:

\text{Let}\\\\x-\text{a volume of}\ 20\%\ \text{saline solution}\\\\y-\text{a volume of }\ 50\%\ \text{saline solution}\\\\p\%=\dfrac{p}{100}\\\\20\%=\dfrac{20}{100}=\dfrac{2}{10}=0.2\\\\50\%=\dfrac{50}{100}=\dfrac{5}{10}=0.5\\\\0.2x-\text{a volume of salt in }\ 20\%\ \text{saline solution}\\0.5y-\text{a volume of salt in }\ 50\%\ \text{saline solution}\\\\60mL-\text{a volume of}\ 28\%\ \text{saline solution}\\\\28\%=\dfrac{28}{100}=0.28

0.28\cdot60mL=16.8mL-\text{a volume of salt in }\ 28\%\ \text{saline solution}\\\\\text{Equations}\\\\(1)\qquad x+y=60\\(2)\qquad0.2x+0.5y=16.8

\left\{\begin{array}{ccc}x+y=60\\0.2x+0.5y=168&\text{multiply both sides by 10}\end{array}\right\\\left\{\begin{array}{ccc}x+y=60&\text{multiply both sides by (-2)}\\2x+5y=168}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}-2x-2y=-120\\2x+5y=168}\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad3y=48\qquad\text{divide both sides by 3}\\.\qquad y=16

\text{Put the value of}\ y\ \text{to the first equation:}\\\\x+16=60\qquad\text{subtract 16 from both sides}\\x=44

6 0
4 years ago
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