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Nataly_w [17]
3 years ago
13

What volume of 0.500 M zinc nitrate contains 2.94 g of zinc nitrate?

Chemistry
1 answer:
krek1111 [17]3 years ago
7 0
24518 is the answer to your queston
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which occupies a larger volume, 600 g of water (with a density of 0.995 g/cm3 ) or 600 g of lead (with a density of 11.35 g/cm3
Montano1993 [528]
Water will occupy larger volume
5 0
3 years ago
Determine the mass of 1 mol of gold (Au), of aluminum chloride (AlCl3), and of glucose (C6H12O6)
sleet_krkn [62]

Answer: Mass of 1 mol of gold (Au) is 197g , of aluminum chloride(AlCl_3) is 133.5 g and of glucose (C_6H_{12}O_6) is 180 g

Explanation:  Mass of 1 mole of a substance is called its molar mass.

a) 1 mole of Au contains 6.023\times 10^{23}atoms and weigh 197g.

b) 1 mole of AlCl_3 will weigh=1\times {\text {mass of 1 mole of Aluminium}}+3\times {\text {mass of 1 mole of chlorine atom}}

Thus mass of 1 mole of AlCl_3 will be =1moles\times {27g/mol}+3moles\times {35.5g/mol}=133.5g

c) 1 mole of C_6H_{12}O_6 will weigh=6\times {\text {mass of 1 mole of carbon}}+12\times {\text {mass of 1 mole of hydrogen atom}}+6\times {\text {mass of 1 mole of oxygen}}

1 mole of C_6H_{12}O_6 will weigh=6\times 12g/mol+12\times 1g/mol+6\times 16g/mol=180g


5 0
3 years ago
What is the relationship between frequency and energy? (Director Inverse)
mars1129 [50]

Answer: The energy of a wave is directly proportional to its frequency, but inversely proportional to its wavelength. In other words, the greater the energy, the larger the frequency and the shorter (smaller) the wavelength.

Explanation:

4 0
3 years ago
How can you tell states of matter apart?​
sergejj [24]

You can tell them apart from its form.

If it forms loose and needs like a container to hold it up, that's liquid.

If it is hard and stable, definitely a solid.

And if you barely see something very loose (and normally goes up), that's liquid.

8 0
3 years ago
Read 2 more answers
A chemist carefully measures the amount of heat needed to raise the temperature of a 0.47 kg sample of C6H7N from 30.5 degrees C
Vadim26 [7]

Answer:

Mass of the sample = m = 894.0 g

Initial temperature of the sample = T_i=-5.8^oC=267.35 KT

i

=−5.8

o

C=267.35K

Final temperature of the sample = T_f=17.5^oC=290.65 KT

f

=17.5

o

C=290.65K

Change in temperature = \Delta T=(T_f-T_i)ΔT=(T

f

−T

i

)

Specific heat capacity of the substance = c

Heat required to raise the temperature of a 894.0g sample = Q

Q = 4.90kJ = 4900 J

Q=m\times c\times \Delta T=m\times c\times (T_f-T_i)Q=m×c×ΔT=m×c×(T

f

−T

i

)

4900 J=894.0 g\times c\times (290.65 K-267.35 K)4900J=894.0g×c×(290.65K−267.35K)

c=0.23523 J/g K\approx 0.235 J/g Kc=0.23523J/gK≈0.235J/gK

The chemist will report the specific heat capacity of the substance as 0.235 J/g K.

5 0
3 years ago
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