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Mice21 [21]
3 years ago
11

Help Please and answer correctly otherwise, I can report the answer and your account so make sure the answer is correct BUT PLEA

SE HELP ME I AM BEGGING. BRAINLY PLEASE PUT ME AT THE TOP OF THE PAGE PLEASE.

Chemistry
1 answer:
Alina [70]3 years ago
8 0

Answer:

G

Explanation:

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State Boyle's, Charles's, and Gay-Lussac's laws using sentences, then equations
dimaraw [331]

Explanation:

1. Boyle's Law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}      (At constant temperature and number of moles)

P_1\times {V_1}=P_2\times V_2

2. Charles' Law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

V\propto T    (At constant pressure and number of moles

\frac{V_1}{T_1}=\frac{V_2}{T_2}

3. Gay Lussac's Law states that tempertaure is directly proportional to the pressure of the gas at constant volume and number of moles of gas

P\propto T    (At constant volume and number of moles)

\frac{V_1}{n_1}=\frac{V_2}{n_2}

7 0
3 years ago
4 Infer is the reaction below possible?
mars1129 [50]
No, because hydrogen isn’t brought out of the equation
8 0
3 years ago
A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions
Mars2501 [29]

Answer:

Qm  = -55.8Kj/mole

Explanation:

NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)

Qm = (mc∆T)water /moles acid

Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)

=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)

=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)

ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃

= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*

Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.

4 0
3 years ago
Read 2 more answers
Consider the reaction below.
Bond [772]
It decreases i believe
7 0
3 years ago
Is the normal resting position of an object.
NemiM [27]

Answer:

being stationary relative to a particular frame of reference or another object; when the position of a body with respect to its surroundings does not change with time it is said to be at rest

Explanation:

4 0
3 years ago
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