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3 years ago

If one pound is the same as 454 grams, then convert the mass of 78 grams to pounds.

Chemistry

1 answer:

LiRa [457]3 years ago

4 0

Answer:

0.17 lb

Explanation:

78 g * (1 lb/454 g)=0.17 lb

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What tool did scientist use to first study the Earth’s interior

IgorC [24]

Seismic waves hope this helps.

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3 years ago

Which has more atoms, 12.0 mol of C or 12.0 mol of Ca?

mr_godi [17]

Answer:

they are equal.

Explanation:

1 mol = 6.022 × 10^23 (Avogadro's constant), which is the number of atoms in 1 mol of any element. Doesn't matter what their atomic mass is, although, of course, 1 mol of carbon weighs less than 1 mol of calcium, but its because their mass is different, but the point is, in 1 mol of any element there is 6.03*10^23 atoms

This is like saying, what weighs more, 10 kg of feathers or 10 kg of metal

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2 years ago

Nitric acid, a major industrial and laboratory acid, is produced commercially by the multi-step Ostwald process, which begins wi

Marizza181 [45]

Answer: 3.178 Kg of ammonia must be used to produce 7.839 kg of nitric acid and The equations are not balanced. Explanation: Ostwald process is a multi-step for manufacturing Nitric Acid. First, We must check if our equations are balanced or not, by checking the amount of each element before and after the arrow. Step 1: NH3(g) + O2(g) ⟶ NO(g) + H2O(l) Step 2: NO(g) + O2(g) ⟶ NO2(g) Step 3: NO2(g) + H2O(l) ⟶ HNO3(l) + NO(g) For example in Step 1, the amount of Hydrogen atoms are different on both sides. On left side we can count 3 hydrogen atoms while on the right side we can count only 2 hydrogen atoms. In the Step 2, the amount of Oxygen atoms are different on both sides too. On the Left side we can count 3 oxygen atoms while on the right side we can count only 2 oxygen atoms and in step 3 the amount of Nitrogen atoms are different on both sides too. On the left side we can count 1 nitrogen atom while on the right side we can count 2 nitrogen atoms. So, Let´s balance equations by inspection, just adding coefficients to each compound, until the amount of atoms are equal on both sides. Step 1: 4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l) Step 2: 2NO(g) + O2(g) ⟶ 2NO2(g) Step 3: 3NO2(g) + H2O(l) ⟶ 2HNO3(l) + NO(g) Second we gather the information what we are going to use in our calculations. Final Mass of HNO3 = 7,839Kg and Molecular Weigth = 63,01g/mol NH3 Molecular Weight= 17,031 g/mol Third we start discoverying the amount of NH3 that reacted completed to generate 7,839Kg of HNO3, using the giving equation and its respective molecular weights. 7,839Kg of HNO3 x 1000g / 1Kg x 1mol HNO3 / 63,01g HNO3 x 3mol NO2 / 2 mol HNO3 x 2 mol NO/ 2 mol NO2 x 4 mol NH3/4 mol NO x 17,031g NH3/1 mol NH3 x 1 Kg / 1000g= 3,178 Kg NH3 The amount of NH3 that is required to produce 7,839 Kg of HNO3 is 3,178 Kg NH3

7 0

3 years ago

calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?

inessss [21]

Answer:

$\Delta E=E_{final}-E_{initial}$

$\Delta E=-1312[\frac{1}{(n_f^2)}-\frac {1}{(n_i^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{3^2)}-\frac {1}{(1^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{(9)}-\frac {1}{(1 )}]KJ mol^{-1}$

$\Delta E=-1312[0.111-1]KJ mol^{-1}$

$\Delta E=1166 KJ mol^{-1}$

$\frac{=1166,000 \mathrm{J}}{6.022 \times 10^{23} \text { photons }}$

$=193623 \times 10^{-23} \frac {J}{photon}$

$\Delta E=1.93623 \times 10^{-18} \frac {J}{photon}$

$\Delta E=\frac {h\times c}{\lambda} \\\\=\frac {(6.626\times 10^{-34} J s \times 3 \times 10^8 ms^{-1})}{\lambda}$

h is planck's constant

c is the speed of light

λ is the wavelength of light

$\lambda =\frac {h\times c}{\Delta E}\\\\=\frac {(6.626\times10^{-34} J s\times3 \times 10^8 ms^{-1})}{(1.93623\times10^{-18} J/photon)}$

Wavelength

$\lambda =10.3 \times 10^{-8} m \times \frac {(10^9 nm)}{1m} =103 nm (Answer)$

Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is 103 nm.

7 0

3 years ago

3. A compound was analyzed and found to contain 9.8 g of nitrogen, 0.70 g of

Nesterboy [21]

Answer:

HNO₃

Explanation:

Data given

Nitrogen = 9.8 g

Hydrogen = 0.70 g

Oxygen = 33.6 g

Empirical formula = ?

Solution:

Convert the masses to moles

For Nitrogen

Molar mass of N = 14 g/mol

no. of mole = mass in g / molar mass

Put value in above formula

no. of mole = 9.8 g/ 14 g/mol

no. of mole = 0.7

mole of N = 0.7 mol

For Hydrogen

Molar mass of H = 1 g/mol

no. of mole = mass in g / molar mass

Put value in above formula

no. of mole = 0.70 g/ 1 g/mol

no. of mole = 0.7

mole of H = 0.7 mol

For Oxygen

Molar mass of O = 16 g/mol

no. of mole = mass in g / molar mass

Put value in above formula

no. of mole = 33.6 g / 16 g/mol

no. of mole = 2.1

mole of O = 2.1 mol

Now we have values in moles as below

N = 0.7

H = 0.7

O = 2.1

Divide the all values on the smallest values to get whole number ratio

N = 0.7 / 0.7 = 1

H = 0.7 / 0.7 = 1

O = 2.1 / 0.7 = 3

So all have following values

N = 1

H = 1

O = 3

So the empirical formula will be HNO₃ i.e. all three atoms in simplest small ratio.

6 0

3 years ago