Answer:
Mg will replace Ag in a compound
Explanation:
A single replacement reaction is driven by the position of ions on the activity series.
As a rule of thumb, the position of metal ions on the activity series determines their reactivity.
Metal ions that are above another are more reactive and they will displace those that are lower.
Generally, activity increases as we go up the group.
Mg ions are higher than Ag ions on the series so, Mg will displace Ag from a solution.
Answer:
Explanation:
1. FALL PROTECTION-GENERAL REQUIREMENTS (29 CFR 1926.501) 6,010 VIOLATIONS
2. HAZARD COMMUNICATION (29 CFR 1910.1200). 3,671
3. SCAFFOLDING (29 CFR 1926.451). 2,813
<span>I would say greater than because as you do deeper, the pressure strengthens. If you were in a 10 ft deep pool and you dive all the way to the bottom, the ears usually pop. That's because of the pressure. Whereas if you were to go five feet, your ears wouldn't. It depends on the age of the person.
Hope this helps.</span>
I'm assuming it was to keep the data consistent? The further you are from a heat source the less heat will get to you as the temperature tries to reach equilibrium and the waves start to spread out, so you should keep everything the same distance to get consistent results. I don't have any information so this is just my assumption
The complete observation about adding bulb 3 is the brightness of the bulbs has to do with power which considers both the voltage and the current: less voltage x less current = dimmer bulbs. In circuit A, the voltage is divided across the resistors and the current decreases as resistance increases. In circuit B, the voltage is the same in each parallel section of the circuit and the current through that section of the circuit only depends on the resistor in that section.
<h3>What is power of the circuit?</h3>
The power of the bulb or any resistor is equal to the product of voltage and current flowing through it.
P = VI
Circuit A has bulbs in series while the circuit B has bulbs in parallel.
When bulb 3 added to circuit A, the brightness of all the bulbs dimmed but when bulb 3 (R3) added to circuit B, nothing changed in the brightness of the bulb.
The brightness is depended on the power of the circuit. When both the voltage and current are less, the bulb will be dimmed. In circuit A, series resistors divide the voltage across them. In circuit B, voltage is equal for all the resistors.
Thus, the last option is correct.
Learn more about power.
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