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hoa [83]
3 years ago
11

How many kilometers are in a 5.44 mile race

Chemistry
2 answers:
OLga [1]3 years ago
7 0

Answer: 8. 75 kilometers

Explanation:  5.44x 1.609344= 8.7548 Kilometers

german3 years ago
4 0

Answer: 8.754831

Explanation:

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Calculate the mass of H2OH2O produced by metabolism of 2.4 kgkg of fat, assuming the fat consists entirely of tristearin (C57H11
otez555 [7]

Answer:

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

Explanation:

2C_{57}H_{110}O_6+163O_2\rightarrow 114CO_2+110H_2O

Mass of fat = 2.4 kg = 2.4 × 1000 g = 2400 g

1 kg = 1000 g

Molar mass of fat = M

M = 57 × 12 g/mol + 110 × 1 g/mol+ 6 × 16 g/mol = 890 g/mol[/tex]

Moles of fat = \frac{2400 g}{890 g/mol}=2.6966 mol

According to reaction , 2 moles of fat gives 110 moles of water. Then 2.6966 moles of fat will give  ;

\frac{110}{2}\times 2.6966 mol=148.31 mol of water

Mass of 148.31 moles of water ;

148.31 mol × 18 g/mol = 2,669.58 g

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

8 0
3 years ago
What volume in liters of carbon monoxide will be required to produce 18.9 L of nitrogen in the reaction below
mina [271]

Answer:

37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.

Explanation:

Equation for the reaction:

2 CO + 2 NO ------> N2 + 2 CO2

2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen

At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.

So therefore, we can say:

2 * 22.4 L of CO produces  22.4 L of N2

44.8 L of CO produces 22.4 L of N2

Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:

44.8 L of CO = 22.4 L of N

x L = 18.9 L

x L = 18.9 * 44.8 / 22.4

x L = 18.9 * 2

x = 37.8 L

The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L

8 0
3 years ago
H2, N2, O2 molecules. . . A)must be polar. must be nonpolar. . B)can be polar or nonpolar depending on geometric . C)configurati
LenKa [72]

Answer: The correct option is A.

Explanation: The given molecules are the molecules of same element.

These molecules are considered as diatomic species.

Polar molecules are the molecules in which some polarity is present in the bond. These molecules are formed when there is some difference in the electronegativities of the elements. Example: HCl

Non-polar molecules are the molecules where no polarity is present in the bond. These molecules are formed when there is no difference in the electronegativities of the elements. Example: H_2, O_2

The given molecules are non-polar in nature.

Hence, these molecules must be non-polar. So, the correct option is A.

6 0
2 years ago
Read 2 more answers
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base
tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

HN_3 + OH- ---> N^-_{3} + H_2O

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of HN_3 = \dfrac{0.001755}{0.0383} = 0.0458 M

Concentration of N^{-}_3 = \dfrac{ 0.001995 }{0.0383} = 0.0521 M

GIven that :

Ka = 1.9 x 10^{-5}

Thus; it's pKa = 4.72

pH =4.72 +  log(\dfrac{ \ 0.0521}{0.0458})

pH =4.72 + log(1.1376)

pH =4.72 + 0.05598

pH =4.77598

pH ≅ 4.80

3 0
3 years ago
2. What is the concentration of hydrochloric acid?
nika2105 [10]
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<h2><em><u>Please</u></em><em><u> </u></em><em><u>mark</u></em><em><u> </u></em><em><u>may</u></em><em><u> </u></em><em><u>be</u></em><em><u> </u></em><em><u>brainliest</u></em><em><u> </u></em><em><u>answer</u></em><em><u>.</u></em><em><u> </u></em></h2>

3 0
1 year ago
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