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2 years ago

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What is the specific heat of a substance if a mass of 10.0 kg increases in temperature from 10.0°c to 70.0°c when 2,520 j of hea

t is applied? use q equals m c subscript p delta t.. 0.00420 j/(gi°c) 0.00661 j/(gi°c) 238 j/(gi°c) 252 j/(gi°c)

2 answers:

Oliga [24]2 years ago

8 0

Answer:

4.2 j/kg-C = .0042 j / (g-C)

Explanation:

Specific heat units are j / kg-C

2520 J / [ (10 kg)(70-10 C)] = 4.2 j / kg-C

LekaFEV [45]2 years ago

5 0

Answer:

0.00420

Explanation:

The equation is Q=mc(T(final)-T(initial), where c is the specific heat, Q is heat supplied, m is mass, T(final) is final temperature and T(initial) is initial temperature (you'll see this written as delta T, which means change in temperature).

2520 = Q

m = 10.0kg; answer choices are in g, not kg, so multiply by 1000 to get m in g; m = 10000 g

Plug in the values you have and solve for c.

2520=(10000)(c)(70-10)

2520=600000c

c=0.0042 j/(gc)

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What is an example of entropy from everyday life?

Deffense [45]

A Campfire is an example, the solid wood becomes ash.

8 0

3 years ago

Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0

3 years ago

In Chemistry, to be classified as an organic substance, a substance must contain

Natasha_Volkova [10]

I'm going on a limb here, but Carbon is a definite. <span />

4 0

2 years ago

Bromine has two isotopes, Br79 and Br81. The isotopes occur in a 50:50 (1:1) ratio. Given that the mass spectrum of bromine cont

AlekseyPX

Answer:

There will be 3 peaks.

Relative height of the atomic peaks would be; 158, 160 and 162

Explanation:

We are told that Bromine has two isotopes namely 79Br and 81Br in a 1 : 1 ratio (50 : 50).

This means that a compound which contains 1 bromine atom will have two peaks in the molecular ion region but it depends on which bromine isotope is contained in the molecular ion.

Thus;

Relative height of atomic peaks is given by;

m/z = 79Br¯ 79Br+ = 158

79Br¯ 81Br+ = 160

81Br¯ 81Br+ = 162

7 0

3 years ago

We make a basic solution by mixing 50. mL of 0.10 M NaOH and 50. mL of 0.10 M Ca(OH)2. It requires 250 mL of an HCl solution to

andreyandreev [35.5K]

<u>Answer:</u> The correct answer is Option 5.

<u>Explanation:</u>

To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

$M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of the NaOH.

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of the $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.10M\\V_1=50mL\\n_2=2\\M_2=0.1\\V_2=50mL$

Putting all the values in above equation, we get:

$M=\frac{(1\times 0.1\times 50)+(2\times 0.1\times 50)}{50+50}\\\\M=0.15M$

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base.

We are given:

$n_1=1\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.15M\\V_2=100mL$

Putting values in above equation, we get:

$1\times M_1\times 250=1\times 0.15\times 100\\\\M_1=0.06M$

Hence, the correct answer is Option 5.

7 0

3 years ago