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3 years ago

The temperature of a 10g sample of iron was raised by 25.4ᵒC with the addition of 114 J

Chemistry

1 answer:

Sidana [21]3 years ago

5 0

Answer:

28956J

Explanation:

Specific heat capacity is equal to mass × heat capacity × temperature change

Shc = mCtheta

= 10 × 114 ×25.4

= 28956J

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A chemist has 60ml of a 3M NaCl solution. He needs to dilute it so the Molarity is 1M NaCl. How much water does he need to add?

REY [17]

Answer:

80 ml the right answer

Explanation:

Hope it helps you dude 

8 0

3 years ago

Помогите срочно сделать по химии промежуточную отестацию за 8 класс

il63 [147K]

Answer:

This is site for English speakers. Этот сайт на английском, поэтому вопрос могут удалить

Explanation:

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4. 3)

5. 4)

6. 2)

7. 1)

8. 4)

9. 3)

10. 3)

11. SO3, H2SO4, Na2SO4

12.

A) оксид меди (II) 2) CuO

Б) хлорид меди(II) 4) CuCl2

В) сульфит меди (II) 3) CuSO 3

Г) гидроксид меди (II) 1) Cu(OH)2

13.

1. Fe+HCl= б) FeCl 2 +H 2

2.Fe+O2= в) Fe 3 O 4

3. Fe(OH) 3 = г)Fe 2 O 3 +H 2O

4. FeCl 2 +NaOH= а) Fe(OH) 2 +NaCl

14. 2Ca + O2 = 2CaO

CaO + H2O = Ca(OH)2

Ca(OH)2 + 2HCl = CaCl2 + 2H2O

6 0

3 years ago

The following information was recorded by a student team working to prepare nickel sulfate. Plan: Prepare NiSO4 by reacting NiO

mixas84 [53]

Answer:

The options e and d are correct.

Explanation:

Mass of NiO = 7.5 g

Moles of NiO = $\frac{7.5 g}{74.69 g/mol}=0.10 mol$

Moles of sulfuric acid = n

Volume of sulfuric acid ,V= 50 mL = 0.050 L

Molarity of sulfuric acid ,M = 6 mol/L

$n=M\time V=6mol/L\times 0.050 L =0.3 mol$

$NiO + H_2SO_4\rightarrow NiSO_4 + H_2O$

According to reaction, 1 mole of NiO reacts with 1 mole of sulfuric acid.

Then 0.10 moles of NiO reacts with :

$\frac{1}{1}\times 0.10 mol/=0.10 mol$ of sulfuric acid.

As we can see that sulfuric acid is in excess amount, so the amount of the product will depend upon amount of NiO.

According to reaction, 1 mole of NiO gives with 1 mole of $NiSO_4$ .

Then 0.10 moles of NiO wil give :

$\frac{1}{1}\times 0.10 mol/=0.10 mol$ of $NiSO_4$ .

Molar mass of $NiSO_4$ = 154.75 g/mol

Mass of 0.10 moles of $NiSO_4$ :

= 154.75 g/mol × 0.10 mol = 15.475 g

Theoretical mass of $NiSO_4$ = 15.475 g

Experimental yield of $NiSO_4$ = 17.4 g

Percentage yield :

$\Yield=\frac{\text{Experimental mass}}{\text{Theoretical mass}}\times 100$

Percentage yield of $NiSO_4$ :

$\Yield=\frac{17.4}{15.475 g}\times 100=112\%$

Moles of $NiSO_4.6H_2O$ = 262.85 g/mol × 0.10 mol = 26.285 g

Experimental yield of $NiSO_4.6H_2O$ = 17.4 g

Percentage yield of $NiSO_4.6H_2O$ :

$\Yield=\frac{17.4}{26.285 g}\times 100=66.2\%$

3 0

3 years ago

Which scientific discipline belongs in the blue box?

Bess [88]

Answer:

PHYSICS.

Explanation:

BLUE BOX CA BE REFERRED AS AN ELECTRONIC DEVICE WHICH IS USED IN THE TELEPHONIC CIRCIUTSTI PASS ON LONG DIATANT SIGNALS.

7 0

3 years ago

Read 2 more answers

At 63.5 C the vapor pressure of H2O is 175 torr and that of ethanol is 400 torr. A solution is made by adding equal masses of H2

xxTIMURxx [149]

Answer:

Moel fraction of ethanol in the solution = 0.28

Vapor pressure of the solution = 238 torr

Mole fraction of ethanol in the vapor = 0.47

Explanation:

Let's use 100 g of each substance as a calculus basis. Knowing that the molar mass of water is 18 g/mol and the molar mass of ethanol is 46 g/mol, the number of moles (n = mass/molar mass) of each one is:

nw = 100/18 = 5.56 mol

ne= 100/46 = 2.17 mol

The total number of moles is 7.73 mol, so the mole fraction of ethanol is

2.17/7.73 = 0.28

The mole fraction of water must be 0.72, so if we assume that the solution is ideal, by the Raoult's law, the solution vapor pressure is the sum of the multiplication of the mole fraction by the vapor pressure of each substance, thus:

P = 0.28*400 + 0.72*175

P = 238 torr

The partial pressure of each substance can be found by the multiplication of the molar fraction by the vapor pressure, thus:

Pw = 0.72*175 = 126 torr

Pe = 0.28*400 = 112 torr

To know the number of moles that is vaporized above the solution, we may use the ideal gas law:

PV = nRT

P/n = RT/V

R is the gas constant, T is the temperature and V is the volume, so they are the same for both water and ethanol, thus

Pw/nw = Pe/ne

126/nw = 112/ne

ne = (112/126)*nw

ne = 0.89nw

So, the mole fraction of ethanol is:

ne/(ne + nw) = 0.89nw/(0.89nw + nw) = 0.89/1.89 = 0.47

4 0

3 years ago