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Ymorist [56]
3 years ago
5

Which of these is an element?

Chemistry
2 answers:
Brilliant_brown [7]3 years ago
8 0

Answer:

Carbon is the only element.

bearhunter [10]3 years ago
4 0
The answer is A, carbon
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How many kilojoules (kJ) in 373 joules (J)?
Nonamiya [84]

Answer: There are 0.373 kJ in 373 joules.

Explanation:

According to the standard conversion units 1 kJ is equal to 1000 J. This means that 1 J is equal to 0.001 kJ.

Hence, 373 joules is converted into kilojoules as follows.

1 J = 0.001 kJ\\373 J = 373 J \times \frac{0.001 J}{1 kJ}\\= 0.373 kJ

Thus, we can conclude that there are 0.373 kJ in 373 joules.

4 0
3 years ago
Which of these equations are balanced
gogolik [260]

Answer:

Post the equations for help :)

Explanation:

6 0
3 years ago
A buffer solution is prepared by placing 5.86 grams of sodium nitrite and 32.6 mL of a 4.90 M nitrous acid solution into a 500.0
N76 [4]

This question is incomplete, the complete question is;

A buffer solution is prepared by placing 5.86 grams of sodium nitrite and 32.6 mL of a 4.90 M nitrous acid solution into a 500.0 mL volumetric flask and diluting to the calibration mark. If 10.97 mL of a 1.63 M solution of potassium hydroxide is added to the buffer, what is the final pH? The Ka for nitrous acid = 4.6 × 10⁻⁴.

Answer:

the final pH is 3.187

Explanation:

Given the data in the question;

Initial moles of HNO2 = 32.6/1000 × 4.90 = 0.15974 mol

Initial moles of NO2- = mass/molar mass = 5.86/68.995 =  0.0849336 mol

Moles of KOH added = 10.97/1000 × 1.63  = 0.0178811 mol

so

HN02 + KOH → NO2- + H2O

moles of HNO2 = 0.15974 - 0.0178811 = 0.1418589 mol

Moles of NO2- = 0.0849336 + 0.0178811  =  0.1028147 mol

Now,

pH = pka + log( [NO2-]/[HNO2])

pH = -log ka + log( moles of NO2- / moles of HNO2 )

we substitute

pH = -log( 4.6 × 10⁻⁴ ) + log( 0.1028147  / 0.1418589  )

pH = -log( 4.6 × 10⁻⁴ ) + log( 0.724767 )

pH =  3.337242 + (-0.1398 )

pH = 3.187

Therefore, the final pH is 3.187

8 0
3 years ago
Calculate the amount of heat needed to boil 120.g of acetic acid (HCH3CO2), beginning from a temperature of 16.7°C.
evablogger [386]

Answer:

The total amount of heat needed = 72.2116  kJ

Explanation:

Given that ;

the mass of acetic acid = 120.0 g

The initial temperature T_1 = 16.7 °C  = (16.7 + 273.15 ) K = 298.85 K

The standard molar mass of acetic acid = 60.052  g/mol

Thus ; we can determine the number of moles of acetic acid;

number of moles of acetic acid = mass of acetic acid/ molar mass of acetic acid

number of moles of acetic acid = 120.0 g/ 60.052 g/mol

number of moles of acetic acid =  1.998 moles

For acetic acid:

The standard boiling point  T_2 = 118.1 °C = ( 118.1 + 273.15 ) K = 391.25 K

The enthalpy of vaporization of acetic acid \Delta H_{vap} = 23.7 kJ/mol

The heat capacity of acetic acid   c = 2.043  J/g.K

The change in temperature Δ T = T_2 - T_1

Δ T = (391.25 - 289.85)K

Δ T = 101.4 K

The amount of heat needed to bring the liquid acetic acid at 16.7°C to its boiling point is ;

q = mcΔT

From our values above;

q = 120 g ×  2.043  J/g.K × 101.4 K

q = 24859.2  J

q = 24859 /1000 kJ

q = 24.859 kJ

we have earlier calculated our number of moles o f acetic acid to be 1.998 moles;

Thus;

The needed amount of heat = \Delta_{vap} *numbers \ of  \ moles

The needed amount of heat = 23.7 \ kJ/mol * 1.998 \ moles

The needed amount of heat = 47.3526 kJ

Hence;

The total amount of heat needed = 24.859 kJ + 47.3526 kJ

The total amount of heat needed = 72.2116  kJ

4 0
3 years ago
When a 10 ml graduated cylinder is filled to the 10 ml mark, the mass of the water was measured to be 9.955 g. if the density of
Lady_Fox [76]

Given, the density of water is  0.9975 g/ml. Density of water is mass of water per unit volume. Mass of 1 ml of water supposed to be  0.9975 g from density of water. So, mass of 10 ml of water is (0.9975 X 10) g= 9.975 g. From graduated cylinder, mass of 10 ml water is measured to be 9.955 g. So, error for mass of 10 ml water= (9.975-9.955)=0.02 g. Percentage of error for 10 ml water is \frac{0.02}{10} X 100 = 0.2. Error in the mass  for the 10 ml of water is 0.2 %.

5 0
3 years ago
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