Answer is: both reactions are exothermic.
<span>In exothermic reactions, heat is released and enthalpy of reaction is less than zero (as it show second chemical reaction).
According to Le Chatelier's principle when the reaction
is <span>exothermic heat is included as a
product (as it show first chemical reaction).</span></span>
Answer: 5.44×10226.022140857(74)×1023⋅mol−1.
Explanation: So the answer is approx. 0.10⋅mol
Answer:
c. 20.0332 g to 20,0 g
Explanation:
A significant figure is each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit, with the exception of the trailing zeros.
<em>Which of the following examples illustrates a number that is correctly rounded to three significant figures?
</em>
a. 109 526 g to 109 500 g. NO. The rounded number has 4 significant figures: 109 500.
b. 0.03954 g to 0.040 g. NO. The rounded number has 2 significant figures: 0.040.
c. 20.0332 g to 20.0 g. YES. The rounded number has 3 significant figures: 20.0.
d. 04.05438 g to 4.054 g. NO. The rounded number has 4 significant figures: 4.054.
e. 103.692 g to 103.7g. NO. The rounded number has 4 significant figures: 103.7.
Answer:
87.5 mi/hr
Explanation:
Because a = Δv / Δt (a = vf - vi/ Δt), we need to find the acceleration first to know the change in velocity so we can determine the final velocity.
vf = 60 mi/hr
vi = 0 mi/hr
Δt = 8 secs
a = vf - vi/ Δt
= 60 mi/hr - 0 mi/hr/ 8 secs
= 60 mi/hr / 8 secs
= 7.5 mi/hr^2
Now that we know the acceleration of the car is 7. 5 mi/hr^2, we can substitute it in the acceleration formula to find the final velocity when the initial velocity is 50 mi/hr after 5 secs.
vi = 50 mi/ hr
Δt = 5 secs
a = 7.5 mi/ hr^2
a = vf - vi/ Δt
7.5 = vf - 50 mi/hr / 5 secs
37.5 = vf - 50
87.5 mi/ hr = vf
Molar mass of CH2NH2COOH - 75
Given mass of CH2NH2COOH - 30
Moles of CH2NH2COOH = Given mass/ Molar mass
moles of CH2NH2COOH = 30/75 = 0.4 mol
One mole of CH2NH2COOH contains 32 gram of oxygen
0.4 mole of CH2NH2COOH will contain = 0.4 × 32= 12.8 g of oxygen
Answer- the mass of oxygen in 30 g of CH2NH2COOH is 12.8 gram!
