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3 years ago

Does it require more force to stop a bus or a bike moving at the same speed

1 answer:

7 0

Well depends how fast they're going if it's a slow speed a bus but at a fast speed a bike because you wanna be careful while stopping

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How many grams of magnesium metal will react completely with 6.3 liters of 4.5 M HCl? Show all of the work needed to solve this

Rufina [12.5K]

Mg (s) + 2 HCl (aq) -> MgCl2 (aq) + H2 (g)

1) Number of mols of HCl, n

n = 6.3 L * 4.5 mol/L = 28.35 mol

2) ratio: 2 mol HCl / 1 mol Mg = 28.35 mol HCl / x mol Mg

x = 28.35 mol HCL * 1 mol Mg / 2 mol HCL = 14.175 mol Mg

3) Convert mol to mass using atomic mass of Mg

14.175 mol Mg * 24.3 g Mg / mol Mg = 344. 45 g

Answer: 344.45 g

5 0

3 years ago

HCl reacts with Barium Hydroxide to produce barium chloride and water. how many ml of a 3.00 M hydrochloric acid solution would

ollegr [7]

THINK OF GUMMY BEARS IN A TEST TUBE AND IN THAT TEST TUBE THEIRS A SUBSTANCE ACID IT EX PLODS LOOK IT UP ON YOU TUBE

3 0

3 years ago

What evidence supports conservation of matter?

Zinaida [17]

Answer:

i dont knoq

Explanation:

sorry that i couldnt help you

7 0

2 years ago

The atomic number of magnesium (mg) is 12. this is also the number of which of the following

Rzqust [24]

The answer is protons.

8 0

2 years ago

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation f

dmitriy555 [2]

Answer: The mass of cryolite produced is 51.48 kg

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For aluminium oxide:

Given mass of aluminium oxide = 12.5 kg = 12500 g (Conversion factor: 1 kg = 1000 g)

Molar mass of aluminium oxide = 101.96 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium oxide}=\frac{12500g}{101.96g/mol}=122.6mol$

For NaOH:

Given mass of NaOH = 55.4 kg = 55400 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaOH}=\frac{55400g}{40g/mol}=1389mol$

For HF:

Given mass of HF = 55.4 kg = 55400 g

Molar mass of HF = 20 g/mol

Putting values in equation 1, we get:

$\text{Moles of HF}=\frac{55400g}{20g/mol}=2770mol$

For the given chemical reaction:

$Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O(g)$

By Stoichiometry of the reaction:

1 mole of aluminium oxide reacts with 6 moles of sodium hydroxide and 12 moles of HF.

So, 122.6 moles of aluminium oxide will react with $(6\times 122.6)=735.6mol$ of sodium hydroxide and $(12\times 122.6)=1471.2mol$ of HF

As, given amount of NaOH and HF is more than the required amount. So, they are considered as an excess reagent.

Thus, aluminium oxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aluminium oxide produces 2 moles of cryolite

So, 122.6 moles of aluminium oxide will produce = $\frac{2}{1}\times 122.6=245.2mol$ of cryolite

Now, calculating the mass of cryolite by using equation 1:

Molar mass of cryolite = 209.94 g/mol

Moles of cryolite = 245.2 mol

Putting values in equation 1, we get:

$245.2mol=\frac{\text{Mass of cryolite}}{209.94g/mol}\\\\\text{Mass of cryolite}=(245.2mol\times 209.94g/mol)=51477.3g$

Converting this into kilograms, we use the conversion factor:

1 kg = 1000 g

So, $51477.3 g\times (\frac{1kg}{1000g})=51.48kg$

Hence, the mass of cryolite produced is 51.48 kg

7 0

2 years ago