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Gemiola [76]
3 years ago
14

What is the most likely evidence of an energy transformation taking place when a skater glides over ice? A.Skating blades become

cold as kinetic energy is transformed into thermal energy B.Skating blades become cold as kinetic energy is transformed into gravitational energy C.Small tracks of water are left on the ice as kinetic energy is transformed into thermal energy D.Small tracks of water are left on the ice as kinetic energy is transformed into gravitational energy
Chemistry
1 answer:
Diano4ka-milaya [45]3 years ago
7 0
The correct answer is C. The weight downward and the friction moving forward creates thermal energy through kinetics and melts a tiny bit of the ice making it slick enough for the skate to slide.
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Pre-laboratory Questions 1. Using LeChâtelier’s principle, determine whether the reactants or products are favored and the direc
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I think the answer is C but don’t quote me on that
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2 years ago
In a hospital laboratory, a 10.0 mL sample of gastric juice (predominantly HCl), obtained several hours after a meal, was titrat
LiRa [457]

Answer: 1.14

Explanation:

HCl+NaOH\rightarrow NaCl+H_2O

To calculate the molarity of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=?\\V_1=10.0mL\\n_2=1\\M_2=0.1M\\V_2=7.2mL

Putting values in above equation, we get:

1\times M_1\times 10.0=1\times 0.1\times 7.2\\\\M_1=0.072M

To calculate pH of gastric juice:

molarity of H^+ = 0.072

pH=-log[H^+]

pH=-log(0.072)=1.14

Thus the pH of the gastric juice is 1.14

3 0
3 years ago
Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o
Sunny_sXe [5.5K]

Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}

Now, we are able to solve for m:

m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} }  = 9.70*10^4 g Na=97 Kg Na=97 000 g Na

7 0
3 years ago
Read 2 more answers
Write the concentration equilibrium constant expression for this reaction.
Akimi4 [234]

In a chemical reaction, the equilibrium constant refers to the value of its reaction quotient at chemical equilibrium, that is, a condition attained by a dynamic chemical system after adequate time has passed, and at which its composition has no measurable capacity to undergo any kind of further modification.  

The given reaction is: HCN (aq) + OH⁻ = CN⁻ (aq) + H2O (l)

The equilibrium constant = product of concentration of products / product of concentration of reactants

(Here, H2O is not considered as its concentration is very high)

So, Keq = [CN⁻] / [HCN] [OH⁻]


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Stationary Front: a front that is not moving. When a warm or cold front stops moving, it becomes a stationary front.
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