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pantera1 [17]
3 years ago
7

What metalic element is sometimes used in flarres and flashbulbs

Chemistry
1 answer:
hjlf3 years ago
7 0
Magnesium bc it burns a bright light when used
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Whatvdo you meant by acid and base​
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3 years ago
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What volume in milliters would give 12.0 grams of NaOH from a 2.0 M NaOH solution?
zavuch27 [327]

Answer:

V = 150 mL

Explanation:

Hello!

In this case, since the molarity of a solution is computed via:

M=\frac{n}{V}

Whereas we are asked to compute the volume:

V=\frac{n}{M}

Thus, we first compute the moles in 12.0 g of NaOH as its molar mass is about 40.0 g/mol:

n=\frac{12.0g}{40.0g/mol}=0.3mol

Thus, at first instance, the volume liters is:

V=\frac{0.3mol}{2.0mol/L} \\\\V=0.15L

Which in milliliters is:

V=0.15L*\frac{1000mL}{1L}\\\\V=150mL

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3 years ago
CER sentence starter
jonny [76]

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2 years ago
You placed 6.35 g of a mixture containing unknown amounts of BaO(s) and MgO(s) in a 3.50-L flask containing CO₂(g) at 30.0°C and
adelina 88 [10]

Mass of BaO in  initial mixture = 3.50g

Explanation:

Let mass of BaO in mixture be x g

mass of MgO in mixture be (6.35 - x) g

Initially CO_2

Volume = 3.50 L

Temp = 303 K

Pressure = 750 torr = 750 / 760 atm

Applying ideal gas equation

PV = nRT

n = PV / RT

(n)_CO_2 = ((750/760)* 3.50) / 0.0821 * 303

(n)_CO_2 = 0.139 mole

Finally; mole of CO_2

n= PV /RT

((245/760) *3.5) / 303* 0.0821

(n)_CO_2 = 0.045 mole

Mole of CO_2 reacted = 0.139 - 0.045

=0.044 mole

BaO + CO_2  BaCO_3

Mgo + CO_2  MgCO_3

moles of CO_2 reacted = ( moles of BaO + moles of MgO)

moles of BaO in mixture = x / 153 mole

moles of MgO in mixture = 6.35 - x mole / 40

Equating,

x/ 153 +6.35/40 = 0.094

= x/153 + 6.35 / 40 - x/40 =0.094

= x (1/40 - 1153) = (6.35/40 - 0.094)

= x * 10.018464

= 0.06475

mass of BaO in mixture = 3.50g

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3 years ago
What can happen to the flame if the gas flow is too great or too much air enters the air vent​
rusak2 [61]
The flame goes out. Or it will just increase in size the more gas
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3 years ago
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