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lesya [120]
3 years ago
11

What information does Jaime need to evaluate the accuracy of his calculation?

Chemistry
2 answers:
tiny-mole [99]3 years ago
8 0

Answer:

A. the accepted value for the density of aluminum

Explanation:

To know if your calculation is accurate or not, you need to know the accepted value. An example is if the accepted value was 15 and your calculation was 14.9, your calculation is accurate because its close to the actual value. If your calculation was 7, you would not be accurate at all because you are far from the actual value. Hope this helps please give brainliest!

Effectus [21]3 years ago
4 0

Answer:

well we can't awnser this we don't have the full information srry♥️✨

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A __________ is created by using a secure hash function to generate a hash value for a message and then encrypting the hash code
Basile [38]

Answer: Digital signature

Explanation:

8 0
3 years ago
The water bottle contains 575 grams of water at 80°C. The water eventually cools to
Jet001 [13]

Answer:

–23000 Calories.

NOTE : The negative sign indicates that heat has been loss to the student back.

Explanation:

The following data were obtained from the question:

Mass (M) of water = 575 g

Initial temperature (T1) = 80 °C

Final temperature (T2) = 40 °C

Heat (Q) transferred =.?

Next, we shall determine the change in temperature (ΔT).

This is illustrated below:

Initial temperature (T1) = 80 °C

Final temperature (T2) = 40 °C

Change in temperature (ΔT) =?

Change in temperature (ΔT) = T2 – T1

Change in temperature (ΔT) = 40 – 80

Change in temperature (ΔT) = –40°C

Finally, we shall determine the heat transferred. This can be obtained as follow:

Mass (M) of water = 575 g

Change in temperature (ΔT) = –40°C

Specific heat capacity (C) of water = 1 Cal/g°C

Heat (Q) transferred =.?

Q = MCΔT

Q = 575 × 1 × –40

Q = –23000 Calories

NOTE : The negative sign indicates that heat has been loss to the student back.

8 0
3 years ago
1718 l of a 0.3556-m c3h7oh solution is diluted to a concentration of 0.1222 m, what is the volume of the resulting solution
Agata [3.3K]
Answer : The volume of the resulting solution will be 0.4999 L.

Explanation : As the two molar concentration of the C_{3} H_{7}OH solution are given and one of the volume concentration needs to be found.

So, according to the formula :-  m_{1}V_{1} = m_{2}V_{2}

And here if we consider V_{1} as 0.1718 L moles and m_{1}  as 0.3556 moles then we need to find V_{2} as m_{2}  is given as 0.1222 moles.
Hence on solving the formula we get,  

 V_{2} = (0.1718 X 0.3556) / 0.1222 = 0.4999 L

 


5 0
3 years ago
A shift of a cup of through
KatRina [158]
What are you asking?
5 0
3 years ago
Read 2 more answers
The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C
artcher [175]

Explanation:

The given data is as follows.

         P_{1} = 100 mm Hg or \frac{100}{760}atm = 0.13157 atm

         T_{1} = 1080 ^{o}C = (1080 + 273) K = 1357 K

         T_{2} = 1220 ^{o}C = (1220 + 273) K = 1493 K

         P_{2} = 600 mm Hg or \frac{600}{760}atm = 0.7895 atm

          R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

                   log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}

            log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]

          log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]

                0.77815 = \frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K

              \Delta H_{vap} = 2.219 \times 10^{5} J/mol

                                   = 2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}

                                    = 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0
3 years ago
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