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Tasya [4]
3 years ago
9

Which element is commonly found in meteorites

Chemistry
2 answers:
MA_775_DIABLO [31]3 years ago
3 0
The element that is commonly found in meteorites is oxygen
Lynna [10]3 years ago
3 0
The element commonly found is oxygen
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Consider the balanced chemical reaction below. When the reaction was carried out, the calculated theoretical yield for carbon di
jeka94

Answer:

Percent Yield = 94.237%

Explanation:

CO = Carbon Dioxide = Molar Mass 28g/mol

C = Carbon = 12g/mol

O = Oxygen = 16g/mol

Theoretical yield = 93.7 grams

Actual yield = 88.3 grams

Percent yield  = (actual yield /theoretical yield )x100

Percent Yield = (88.3/93.7)x100

Percent Yield = 94.237%

8 0
3 years ago
Read 2 more answers
A single serving bag of snack chips contains 65.0 Cal. Assuming that all of the energy from eating these chips goes toward keepi
AveGali [126]

Answer:

= 62.1 hours

Explanation:

Energy provide by the serving is 65 cal

= 65 cal  × 4.184 Kj = 271.96 kJ

271.96 KJ = 271960 J

Energy required for 1minute of energy

= 73 x 1

= 73 J/min

So, 271960 joules will be required for 271960 heart beat

Minutes = 271960 / 73

= 3593.94 minutes  

Time in hours = 3725.429 / 60

= 62.1 hours

5 0
4 years ago
What are the products in the chemical equation for the overall process of photosynthesis?
Alik [6]
Glucose and a plants and ur welcome
8 0
3 years ago
At 25.0°c, a solution has a concentration of 3.179 m and a density of 1.260 g/ml. the density of the solution at 50.0°c is 1.249
oksano4ka [1.4K]

Answer: -

3.151 M

Explanation: -

Let the volume of the solution be 1000 mL.

At 25.0 °C, Density = 1.260 g/ mL

Mass of the solution = Density x volume

= 1.260 g / mL x 1000 mL

= 1260 g

At 25.0 °C, the molarity = 3.179 M

Number of moles present per 1000 mL = 3.179 mol

Strength of the solution in g / mol

= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)

Now at 50.0 °C

The density is 1.249 g/ mL

Mass of the solution = density x volume = 1.249 g / mL x 1000 mL

= 1249 g.

Number of moles present in 1249 g = Mass of the solution / Strength in g /mol

= \frac{1249 g}{396.35 g/mol}

= 3.151 moles.

So 3.151 moles is present in 1000 mL at 50.0 °C

Molarity at 50.0 °C = 3.151 M

7 0
3 years ago
A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 m
Lana71 [14]

Answer:

Oxide of M is M_2O_3 and sulfate of M_2(SO_4)_3

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

PV=nRT

n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol

Moles of hydrogen gas produced = 0.01225 mol

2M+2xHCl\rightarrow 2MCl_x+xH_2

Moles of metal =\frac{0.225 g}{27.0 g/mol}=8.3333 mol

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

\frac{8.3333}{0.01225 mol}=\frac{2}{x}

x = 2.9 ≈ 3

2M+6HCl\rightarrow 2MCl_3+3H_2

MCl_3\rightarrow M^{3+}+Cl^-

Formulas for the oxide and sulfate of M will be:

Oxide of M is M_2O_3 and sulfate of M_2(SO_4)_3.

3 0
4 years ago
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