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ruslelena [56]
2 years ago
7

How long would it take Jesse, with an acceleration of -2.50 m/s², to bring his bicycle, with an initial velocity of 13.5 m/s, to

a complete stop?
Chemistry
1 answer:
lbvjy [14]2 years ago
6 0

Answer:

i am so sorry. i do not have a answer but i am trying to find questioms i can answer

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a piece of food is burned in a calorimeter that contains 200.0g of water. If the temperature of the water rose from 65.0°C to 83
Flauer [41]

Answer: 15062.4 Joules

Explanation:

The quantity of heat energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = ?

Mass of food = 200.0g

C = 4.184 j/g°C

Φ = (Final temperature - Initial temperature)

= 83.0°C - 65.0°C = 18°C

Then, Q = MCΦ

Q = 200.0g x 4.184 j/g°C x 18°C

Q = 15062.4 J

Thus, 15062.4 joules of heat energy was contained in the food.

4 0
3 years ago
Atomic mass is typically expressed in
Andre45 [30]

Answer:

amu

Explanation:

Please mark as brainleist

8 0
2 years ago
Read 2 more answers
The theoretical yield of Cl2 from certain starting amounts of MnO2 and HCl was calculated as 60.25 g and 65.02 g, respectively.
user100 [1]

Answer:

c    43.38 g

Explanation:

The reaction  between MnO2 and HCl can be represented by the following balanced equation:

MnO2 + HCl ---> Cl2 + MnCl2 + H2O

From the balanced equation, the theoretically required molar ratio of MnO2 to HCl is 1:1, therefore the yields would have been expected to be equal.  

For the fact that HCl  gives a higher yield (65.02g) than MnO2 (60.25g) according to the problem statement, HCl should be in excess,  while the limiting reagent should be MnO2 .  

Thus, the theoretical yield of Cl2 will be  60.25 g.

By definition, the percentage yield is given by

% Yield = (Actual Yield) / (Theoretical Yield),  

This can be simplified to

Actual Yield = % Yield * Theoretical Yield

Plugging in the given values we have

Actual Yield = 72% *  60.25 = 43.38 g

5 0
3 years ago
What changes happens when you run ?
tekilochka [14]

Answer:

more detail please

Explanation:

6 0
2 years ago
Read 2 more answers
Which of the following buffers will be most effective at pH 9.25? Group of answer choices a mixture of 1.0 M HC2H3O2 and 1.0 M N
ludmilkaskok [199]

Answer:

The most effective buffer at pH 9.25 will be  a mixture of 1.0 M NH3 and 1.0 M NH4Cl

Explanation:

Step 1: Data given

pH of a buffer = pKa + log ([A-]/[Ha])

a mixture of 1.0 M HC2H3O2 and 1.0 M NaC2H3O2 (Ka for acetic acid = 1.8 x 10-5)

pH = -log( 1.8 * 10^-5) + log (1/1)

pH = -log( 1.8 * 10^-5)

pH = 4.74

a mixture of 1.0 M NaCN and 1.0 M KCN (Ka for HCN = 4.9 x 10-10)

pH = -log( 4.9 * 10^-10) + log (1/1)

pH = -log( 1.8 * 10^-5)

pH = 9.30

a mixture of 1.0 M HCl and 1.0 M NaCl

The solution made from NaCl and HCl will NOT act as a buffer.

HCl is a strong acid while NaCl is salt of strong acid and strong base which do not from buffer solutions hence due to HCl PH is less than 7.

a mixture of 1.0 M NH3 and 1.0 M NH4Cl (Kb for ammonia = 1.76 x 10^-5)

Ka * Kb = 1*10^-14

Ka = 10^-14 / 1.76*10^-5

Ka = 5.68*10^-10

pH = -log( 5.68*10^-10) + log (1/1)

pH = -log( 5.68*10^-10)

pH = 9.25

The most effective buffer at pH 9.25 will be  a mixture of 1.0 M NH3 and 1.0 M NH4Cl

8 0
2 years ago
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