Answer:
are characterized by one substance gaining an electron while another substance loses an electron
Explanation:
Redox reaction in chemical reaction in which one substance gaining an electron while another substance loses an electron. This means that one element is oxidize by losing an electron while the other is reduced by gaining an electron. The one oxidized is called reducing agent while the one reduced is called oxidizing agent.
A: The positive charge if the protons in the nucleus equals the negative charge in the electron cloud.
Protons are positive, electrons are negative, and neutrons have no charge/are neutral
Answer:
What are Newton's 1st 2nd and 3rd laws of motion?
They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. More precisely, the first law defines the force qualitatively, the second law offers a quantitative measure of the force, and the third asserts that a single isolated force doesn't exist.
Explanation:
They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. More precisely, the first law defines the force qualitatively, the second law offers a quantitative measure of the force, and the third asserts that a single isolated force doesn't exist.
TheThe distance they have covered for trip is 219Km.
<h3>What do you mean by uniformly accelereted motion?</h3>
When an object is traveling in a straight line with an increase in velocity at equal intervals of time.
The total time for the trip is
T→ t1+ 22 min = t1+ 0.367 h ,where t1 is the time spent traveling at
V1= 89.5 km/ h .
the distance traveled is ∆x = V1t1=Vavg T
after applying value and calculating it we get
t1= 2.44h for a total time of
t total we get 2.81 h .
∆x =V1T1 = VavgTtotal
∆x = 77×2.81= 219 Km
to learn more about Uniform accelereted motion click here brainly.com/question/12920060
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Heat is added to the mass as follows:
Q_-14-0 = mC_iΔT = 39*2.06*(0--14) = 39*2.06*14 = 1124.76 J
Q_0 = mC_f = 39*334 = 13026 J
Q_0-100 = mC_wΔT = 39*4.18*100 = 16302 J
Q_100 = mC_v = 39*2230 = 86970 J
Q_100-108 = mC_sΔT = 39*2.03*(108-100) = 39*2.03*8 = 633.36J
Q = Summation of all the heats added = 1124.76+13026+16302+86970+633.36 = 118056.12 J ≈ 118.06 kJ
