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ratelena [41]
3 years ago
9

Balance the following equation _NaOH (aq) + _HNO3 (aq)> _NaNO3 (aq)+ _H2O (1)

Chemistry
1 answer:
Vlada [557]3 years ago
6 0
Its already balanced, no balancing required
You might be interested in
X(g)+4Y(g)→2Z(g), ΔH∘=−75.0 kJ Before the reaction, the volume of the gaseous mixture was 5.00 L. After the reaction, the volume
vlabodo [156]

Isobaric transition, first law: <span>H=ΔU+w</span> for a gas expansion: <span>w=<span>P<span>ext</span></span>∗ΔV</span> to convert to joules, you need the gas constants. R = 0.08206 L atm/mol*K, R=8.314 J/mol*K <span>w=<span>P<span>ext</span></span>∗ΔV∗<span><span>8.314 J/mol∗K</span><span>0.08206 L atm/mol∗K</span></span></span> <span>ΔU=ΔH−[<span>P<span>ext</span></span>∗ΔV∗<span><span>8.314 J/mol∗K</span><span>0.08206 L atm/mol∗K</span></span>]</span> <span>ΔU=−75000 J−[(43.0atm)∗(2−5)L∗<span><span>8.314 J</span><span>0.08206 L atm</span></span>]</span> Then you need to convert to kJ. by the way U=E, internal energy.
7 0
2 years ago
A 237g sample of molybdnum metal is heated to 100.1 0C and then dropped into an insulated cup containing 244 g of water at 10.0
miv72 [106K]

Answer:

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

Explanation:

We consider the system formed by the molybdenum metal and water as our system, a control mass inside an insulated cup, that is, a container that avoids any energy and mass interactions between system and surroundings.

From statement we notice that metal is cooled down whereas water is heated. According to the First Law of Thermodynamics, we know that:

Q_{metal} - Q_{water} = 0

Q_{metal} = Q_{water}

Where:

Q_{water} - Heat received by water, measured in joules.

Q_{metal} - Heat released by metal, measured in joules.

Now we expand this identity by definition of sensible heat:

m_{metal}\cdot c_{metal}\cdot (T_{m,o}-T) = m_{water}\cdot c_{water}\cdot (T-T_{w,o})

The specific heat of the metal is cleared within equation above:

c_{metal} = \frac{m_{water}\cdot c_{water}\cdot (T-T_{w,o})}{m_{metal}\cdot (T_{m,o}-T)}

If we know that m_{water} = 0.237\,kg, m_{metal} = 0.244\,kg, c_{water} = 4186\,\frac{J}{kg\cdot ^{\circ}C}, T_{w,o} = 10\,^{\circ}C, T_{m,o} = 100.10\,^{\circ}C and T = 15.30\,^{\circ}C, the specific heat of molybdenum is:

c_{metal} = \frac{(0.237\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (15.30\,^{\circ}C-10\,^{\circ}C)}{(0.244\,kg)\cdot (100.10\,^{\circ}C-15.30\,^{\circ}C)}

c_{metal} = 254.119\,\frac{J}{kg\cdot ^{\circ}C}

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

5 0
3 years ago
A
DIA [1.3K]
  • V_1=49.8mL
  • T_1=18°C=291K
  • T_2=83°C=356K

Using Charles law

\\ \sf\longmapsto V_1T_2=V_2T_1

\\ \sf\longmapsto V_2=V_1T_2\div T_1

\\ \sf\longmapsto V_2=\dfrac{49.8(356)}{291}

\\ \sf\longmapsto V_2=\dfrac{17728.8}{291}

\\ \sf\longmapsto V_2=60.9mL

5 0
2 years ago
What is the specific heat of a substance if 1450 calories are required to raise the temperature of a 240g sample by 20℃?
bazaltina [42]

Answer:

6960 J/kg°C

Explanation:

specific heat= mass×specific heat capacity×increase in temperature

specific heat= 0.240×1450×20= 6960 J/kg°C

hope it helps!

5 0
2 years ago
Can you help me with this?
Nikitich [7]

Answer:

hi I'm sorry I can't I just need points

6 0
2 years ago
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