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Vikentia [17]
3 years ago
6

A car travels 350 km in 7 hours. What is the average speed of the car (in km/hr)?

Chemistry
2 answers:
Blababa [14]3 years ago
7 0

Answer:

50 km/hr

Explanation:

just divide the distance by the time (350/7)

anzhelika [568]3 years ago
4 0

Answer:

50 km/hr

Explanation:

Distance = 350 km

Time = 7 hours

Average Speed = ?

Average\:Speed = \frac{Distance}{Time} \\\\Average\: speed = \frac{350\:km}{7\:hr} \\\\A.V = 50\:km/hr

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How do you Calculate the molarity of 250 ml of solution in which 2.7 g of MgCl2 are dissolved
IRINA_888 [86]

Answer:

Molar mass of MgCl2 is 95 g/mol

Mg = 24 g/mol and Cl = 35.5 ×2 = 71 g/mol

moles = mass given/ molar mass

= 2.7/95 = 0.028 mol

volume = 250/1000 = 0.25 dm3 (ml is the same as dm3)

molarity of MgCl2 = moles/volume

= 0.028/0.25

= 0.112 mol/dm3

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¿A qué se debe la coloración de la llama cuando hay un halógeno presente?
Nimfa-mama [501]

Answer:

Green

Explanation:

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8 0
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3 years ago
Read 2 more answers
When methanol, CH 3 OH , is burned in the presence of oxygen gas, O 2 , a large amount of heat energy is released. For this reas
luda_lava [24]

<u>Answer:</u> The mass of methanol that must be burned is 24.34 grams

<u>Explanation:</u>

We are given:

Amount of heat produced = 581 kJ

For the given chemical equation:

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l);\Delta H=-764kJ

By Stoichiometry of the reaction:

When 764 kJ of heat is produced, the amount of methanol reacted is 1 mole

So, when 581 kJ of heat will be produced, the amount of methanol reacted will be = \frac{1}{764}\times 581=0.7605mol

To calculate mass for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of methanol = 0.7605 moles

Molar mass of methanol = 32 g/mol

Putting values in above equation, we get:

0.7605mol=\frac{\text{Mass of methanol}}{32g/mol}\\\\\text{Mass of methanol}=(0.7605mol\times 32g/mol)=24.34g

Hence, the mass of methanol that must be burned is 24.34 grams

4 0
3 years ago
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