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Wittaler [7]
3 years ago
15

Calculate the amount of energy , in Joules, required to raise the temperature of 15.5 g of liquid water from 0.00o C to 75.0 oC.

Chemistry
1 answer:
deff fn [24]3 years ago
8 0

Answer:

10043.225 J

Explanation:

We'll begin by calculating the amount of heat needed to change ice to water since water at 0°C is ice. This is illustrated below:

Mass (m) = 15.5g

Latent heat of fussion of water (L) = 334J/g

Heat (Q1) =..?

Q1 = mL

Q1 = 15.5 x 334

Q1 = 5177 J

Next, we shall calculate the amount of heat needed to raise the temperature of water from 0°C to 75°C.

This is illustrated below:

Mass = 15.5g

Initial temperature (T1) = 0°C

Final temperature (T2) = 75°C

Change in temperature (ΔT) = T2 – T1 = 75 – 0 = 75°C

Specific heat capacity (C) of water = 4.186J/g°C

Heat (Q2) =?

Q2 = MCΔT

Q2 = 15.5 x 4.186 x 75

Q2 = 4866.225 J

The overall heat energy needed is given by:

QT = Q1 + Q2

QT = 5177 + 4866.225

QT = 10043.225 J

Therefore, the amount of energy required is 10043.225 J

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Answer:

3%

Explanation:

Substract the actual error from the final and multiply by 100

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3 years ago
A scuba diver knows that she needs 50.0mol of air for an upcoming dive. What size (volume) tank will she need to fill for this d
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Answer:

1120 L.

Explanation:

Hello!

In this case, as no conditions of pressure of temperature are given for this problem, we can assume that the scuba diver dives at STP (1 atm and 273.15 K), which means that 1 mole of air would occupy a volume of 22.4 L.

In such a way, since she needs 50.0 moles of air, the following ratio is useful to compute the size (volume) of the tank she needs:

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Thereby, we plug in to obtain:

V_2=\frac{22.4L*50.0mol}{1mol}\\\\V_2=1120 L

Best regards!

4 0
3 years ago
When 4.41g of phosphoric acid (H3PO4) react with 9.25g of barium hydroxide, water and insoluble barium phosphate form. [T/I-7] a
AnnZ [28]

Answer:

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

Explanation:

Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.

H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

We will balance it using the trial and error method.

First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

Finally, we will get the balanced equation by multiplying H₂O by 6.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

3 0
3 years ago
Determine the equilibrium constant, Kp, for the following reaction, by using the two reference equations below: 2 NO(g) + O2(g)
klio [65]

Answer:

Kp=3.07x10^6

Explanation:

Hello,

In this case, by knowing the given reference reactions, one could rearrange them as follows:

2 NO(g) \leftrightarrow N_2(g) + O_2(g); Kp_2 = \frac{1}{2.3 x 10^{-19}}=4.35x10^{18}

N_2(g) + 2O_2(g) \leftrightarrow 2NO_2(g);Kp_3=(8.4x10^{-7})^2=7.056x10^{-13}

Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):

2NO(g)+N_2(g)+2O_2\leftrightarrow 2NO_2(g)+N_2+O_2

Consequently, the equilibrium constant is computed as:

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Best regards.

8 0
3 years ago
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Oxidation happens at the anode and reduction happens at the cathode.<span />
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