The answer is Option B , Wash with soap and water.
That is when you are exposed to infectious materials by being accidentally stuck with a used needle , first thing that needs to be done is that wash with soap and water.
When someone used a needle, viruses in their blood, such as hepatitis B or HIV may contaminate it , so it is required to avoid these bacteria when you are exposed to infectious materials by being accidentally stuck with a used needle, wash with soap and water.
Answer:
Mass ratio of sulfur and oxygen in compounds A and B is 3:2 which confirms that the mass ratios in the two compounds are simple multiples of each other
Explanation:
This question seeks to establish/confirm the law of multiple proportions which posits that elements combine to form different substances which are whole number multiples of each other. Best example of this plays out in the formation of several oxides of the same element. Looking at the ratio in which the elements combine in each of the oxides, we can assume that these ratios are simple whole number multiples of each other.
Now back to the question.
In substance A, we have 6 g of sulfur combining with 5.99 g of oxygen
Now, lest us calculate the ratio of the mass of sulfur to that of oxygen = 6g/5.99g = 1
Now let us calculate the mass ratio of sulfur to oxygen in the second compound = 8.6/12.88 = 0.668
Now the ratios in both compounds are 1 to 0.668. 0.668 to fraction is approximately 1/1.5.
So therefore, the ratio we are having would be 1:1/1.5 or 1:0.668
This is same as 1/(1÷1.5) which is 1.5/1 or simply 3/2
This gives a ratio of approximately 1.5 to 1 or 3 to 2
The ratio 3 to 2 indicates that the mass ratios in both com pounds are simple multiples of each other
Answer:
0.19M
Explanation:
2A → A2
Rate constant = 0.0265 M–1min–1
Initial concentration = 2.00 M
Final Concentration = ?
time, t = 180min
The formular relating the parameters is given as;
1 / [A] = kt + 1 / [A]o
1 / [A] = 0.0265 * 180 + (1 / 2)
1 / [A] = 4.77 + 0.5
[A] = 1 / 5.27 = 0.19M
This is a common laboratory experiment called calorimetry which determines the specific heat capacity of the sample metal.
By the Conservation of Energy,
Energy of Metal = Energy of Water
mCmetalΔT = mCwaterΔT, wherein Cwater = 4.187 J/g·°C
(26.5 g)(Cmetal)(98 - 32.5°C) = (150 g)(4.187 J/g·°C)(32.5 - 20°C)
Solving for Cmetal,
<em>Cmetal = 4.523 J/g·°C</em>
