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3 years ago

Can someone help me with these two questions

Chemistry

1 answer:

sergiy2304 [10]3 years ago

6 0

4. The pressure of the inner core is higher than the outer core

5. The coolest layers are farthest from the core

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The gas in a closed container has a pressure of 3.00 x 10² kPa 30 ° C. What will the pressure be if the temperature is lowered t

Rainbow [258]

Answer: 100kPa

Explanation:

P1 = 3.00 x 10² kPa , P2 =?

T1 = 30°C = 30 +273 = 303k

T2 = —172°C = —172 + 273 = 101k

P1/T1 = P2/T2

3 x 10² / 303 = P2 / 101

P2 = (3 x 10² / 303) x 101

P2 = 100kPa

6 0

3 years ago

11. Which is the solubility product expression for Na3PO4(s)?

dlinn [17]

Answer: B

Explanation: the 3 in Na3PO4(s) belong to the sodium atom (Na). so in any of these equations, the 3 would have to be with Na.

A- the 3 is along w the PO4, which would make it part of that bond

C- there is no 3 at all for Na in this choice, making it incorrect

D- Again, the 3 is placed on the other half of the bond

8 0

2 years ago

What are the two types of dichotomous keys. Which is easier to use?

SVEN [57.7K]

I think the two are text and pictorial.

5 0

2 years ago

What type of oxygen balance does trinitrotoluene (TNT, C7H5N3O6) have?

kirza4 [7]

a. Negative

Explanation:

We have to look at its oxidation (burning, explosion) of TNT:

4 C₇H₅N₃O₆ + 21 O₂ → 28 CO₂ + 10 H₂O + 6 N₂ + heat

So trinitrotoluene need more oxygen atoms that is have to be fully oxidized so the oxygen balance will be negative.

An empirical formula with which you calculate the oxygen balance percent (OB %) is:

OB % = (-1600 / molecular wight) × [2X + (Y/2) + M - Z)

were

X = number of carbon atoms

Y = number of hydrogen atoms

M = number of metal atoms

Z = number of oxygen atoms

For TNT:

OB % = (-1600 / 227.13) × [2×7 + (5/2) + 0 - 6)

OB % = -73.97 %

So the oxygen balance percent is negative.

Learn more about:

oxidation of TNT

brainly.com/question/4427298

#learnwithBrainly

8 0

3 years ago

A coffee-cup calorimeter contains 140.0 g of water at 25.1°C . A 124.0-g block of copper metal is heated to 100.4°C by putting i

Kisachek [45]

Answer:

(a) 3347 J; (b) 3043 J; (c) 58 J/K; (d) 35.5 °C

Explanation:

(a) Heat lost by copper

The formula for the heat lost or gained by a substance is

q =mCΔT

ΔT = T₂ - T₁= 30.3 °C - 100.4 °C = -70.1 °C = -70.1 K

q = 124.0 g × 0.385 J·K⁻¹g⁻¹ × (-70.1 K) = -3347 J

The negative sign shows that heat is lost.

The copper block has lost 3347 J.

(b) Heat gained by water

ΔT = 30.3 °C - 25.1 °C = 5.2 °C = 5.2 K

q = 140.0 g × 4.18 J·K⁻¹g⁻¹ × 5.2 K = 3043 J

The water has gained 3043 J.

(c) Heat capacity of calorimeter

Heat lost by Cu = heat gained by water + heat gained by calorimeter

The temperature change for the calorimeter is the same as that for the water.

ΔT = 5.2 K

$\begin{array}{rcl}\text{3347 J} & = & \text{3043 J} + C \times \text{5.2 K}\\\text{304 J} & = & 5.2C \text{ K}\\C & = & \dfrac{\text{304 J}}{\text{5.2 K}}\\\\& = & \text{58 J/K}\\\end{array}$

The heat capacity of the calorimeter is 58 J/K.

(d) Final temperature of water

$\begin{array}{rcl}\text{Heat lost by copper } + \text{Heat gained by water}& = &0 \\\text{Heat lost by copper}& = &-\text{Heat gained by water} \\m_{\text{Cu}}C_{\text{Cu}}\Delta T_{\text{Cu}}& = & -m_{\text{w}}C_{\text{w}}\Delta T_{\text{w}}\\\end{array}\\$

$\begin{array}{rcl}\text{124.0 g} \times \text{0.385 J$\cdot$K$^{-1}$g$^{-1}$}\times \Delta T_{\text{Cu}}& = & -\text{140.0 g} \times 4.18 \text{ J$\cdot$ K$^{-1}$g$^{-1}$}\times \Delta T_{\text{w}}\\\text{47.7 J$\cdot$K$^{-1}$}\times \Delta T_{\text{Cu}}& = &-\text{585 J$\cdot$ K$^{-1}$g}\times \Delta T_{\text{w}}\\\Delta T_{\text{Cu}} & = & -12.26\Delta T_{\text{w}}\\\end{array}$

$\begin{array}{rcl}\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26(\Delta T_{\text{f}} - 30.3\, ^{\circ}\text{C})\\\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26\Delta T_{\text{f}} + 371\, ^{\circ}\text{C}\\13.26\Delta T_{\text{f}} & = & 471\, ^{\circ}\text{C}\\\Delta T_{\text{f}} & = & 35.5\, ^{\circ}\text{C}\\\end{array}$

The final temperature of the water would be 35.5 °C.

7 0

2 years ago