Remembering the equation Q=MCdeltaT where
q=is the amount of heat energy
M=mass
C=specific heat
deltaT= change in temperature
Therefore, using the equation we can substitute values and solve for q.
Q= (15 grams) (0.129 J/(gx°C))(85-22)
Q=(15) ((0.129 J/(gx°C)) (63)
Q=121.9 Joules
The energy needed to raise the temperature of 15 grams of gold from 22 degrees Celsius to 85 degrees Celsius is then 121.9 Joules or 122 Joules (if rounded up).
Ozone which is present in the stratospheric region of atmosphere is helpful for preventing harmful UV rays from reaching the surface of earth. Due to human activity, several compounds (specifically chlorofluorocarbons) are released in atmosphere. Due to inherent chemical stability of these compounds, the remain stable in lower region of atmosphere and slowly diffuse into stratosphere. On reaching the stratosphere, these compounds reacts with ozone and thereby depletes the effective concentration of ozone present in atmosphere. Hence, <span>the Montreal Protocol was signed in 1987 by major countries of the world. This aim of this protocol was to protect the stratospheric ozone layer by phasing out the production and consumption of ozone-depleting substances.</span>
Answer:
I believe the answer is 1m/s²
Explanation:
acceleration= change in speed ÷ time (I hope this is correct)
Answer:
Clear Communication
Explanation:
Clear communication in key when working with a team on a project. Without communication things get messy. Having clear communication with your team mates helps to prevent miscommunication, issues in planning and completing the project, and more. Clear communication can also help you to hear everybody out to come out with the best version of your project and to prevent fighting that would take up time that you could be working.
Answer:
1.60.
Explanation:
- The no. of millimoles of HCl = MV = (0.15 M)(20.0 mL) = 3.0 mmol.
- The no. of millimoles of KOH = MV = (0.10 M)(20.0 mL) = 2.0 mmol.
<em>Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.</em>
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∴ M of remaining HCl [H⁺] remaining = (NV)HCl - (NV)KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.
∵ pH = - log[H⁺]
<em>∴ pH = - log[H⁺] </em>= - log(0.025) = <em>1.602 ≅ 1.60.</em>
