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2 years ago

When 200 grams of water cools from 50.°C to 25°C, the total amount of heat energy released by the water is?

1 answer:

6 0

We will use the formula q = mCΔT. where q = heat energy released. m= mass. C= specific heat of water. ΔT= change in temperature. Specific heat of water has not been given so we will just use the standard value of that which is 4.186J/g °C. q = mCΔT. q = 200 × 4.186 × (50 -25) q = 200 × 4.186 × 25 q = 5000 × 4.186 q = 20, 930 The total amount of energy released is 20, 930 J

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A 600.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution after the addition of 600.0 mL of Na

Leona [35]

Answer: pH=12.69

Explanation:

${\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of HF}=0.20M\times 0.6L=0.12 moles$

$HF\rightarrow H^++F^-$

Initial 0.12 0 0

Eqm 0.12-x x x

$K_a=\frac{[H^+][F^-]}{[HF]}$

$3.5\times 10^{-4}=\frac{x^2}{0.12-x}$

(neglecting small value of x in comparison to 0.12)

$x=4.2\times 10^{-5}$

Moles of $H^+=4.2\times 10^{-5}$

$NaOH\rightarrow Na^++OH^-$

${\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of NaOH}}=0.10M\times 0.6L=0.06 moles$

0.06 moles of NaOH will give 0.06 moles of $[OH^-]$

Now $4.2\times 10^{-5}$ moles of $OH^-$ will be neutralized by $4.2\times 10^{-5}$ moles of $H^+$ and $(0.06-4.2\times 10^{-5})=0.059$ moles of $OH^-$ will be left.

Molarity of $OH^-=\frac{0.059moles}{1.2L}=0.049M$

$pOH=-\log[OH^-]=-\log[0.049]=1.31$

pH = 14 - pOH= 14 - 1.31 = 12.69

5 0

2 years ago

An ion has 24 protons and 19 electrons. Write the symbol for this ion.

Lerok [7]

Answer:

Explanation:

it hs 5 more protons thant electrons, so it has a positive charge of 5

5 0

2 years ago

A 0.500 g sample of tin (Sn) is reacted with oxygen to give 0.534 g of product. What is the empirical formula of the oxide?

REY [17]

Answer:

$Sn_2O$

Explanation:

Hello,

In this case, given that the mass of the product is 0.534 g, we can infer that the percent composition of tin is:

$\%Sn=\frac{0.500g}{0.534g}*100\%\\ \\\%Sn=93.6\%$

Therefore, the percent composition of oxygen is 6.4% for a 100% in total. Thus, with such percents we compute the moles of each element in the oxide:

$n_{Sn}=93.6gSn*\frac{1molSn}{118.8gSn} =0.788molSn\\\\n_O=6.4gO*\frac{1molO}{16gO}=0.4molO$

In such a way, for finding the smallest whole number we divide the moles of both tin and oxygen by the moles of oxygen as the smallest moles:

$Sn:\frac{0.788}{0.4}=2\\ \\O:\frac{0.4}{0.4}=1$

Therefore, the empirical formula is:

$Sn_2O$

Best regards.

8 0

2 years ago

The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s

denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0

2 years ago

The table above summarizes data given to a student to evaluate the type of change that took place when substance X was mixed wit

Tamiku [17]

The question is incomplete, the complete question is;

The table above summarizes data given to a student to evaluate the type of change that took place when substance X was mixed with water. The student claimed that the data did not provide enough evidence to determine whether a chemical or physical change took place and that additional tests were needed. Which of the following identifies the best way to gather evidence to support the type of change that occurred when water and Xwere mixed?

A. Measuring the melting point of the mixture of water and X

B. Adding another substance to the mixture of water and X to see whether a solid forms

C Measuring and comparing the masses of the water, X, and the mixture of water and X

D Measuring the electrical conductivities of X and the mixture of water and X

Answer:

D Measuring the electrical conductivities of X and the mixture of water and X

Explanation:

Unfortunately, I am unable to reproduce the table here. However, from the table, the temperature of the of the mixture of the solid X and water was 101.6°C. This is above the boiling point of water and way below the temperature of the solid X.

This goes a long way to suggest that there was some kind of interaction between the water and X which accounted for the observed temperature of the system of X in water.

The only way we can be able to confirm if X actually dissolved in water is to measure the conductivity of the water. dissolved solids increase the conductivity of water.

6 0

2 years ago