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ollegr [7]
3 years ago
11

When is this going to be the number one question is this a water or liquid​

Chemistry
1 answer:
sergejj [24]3 years ago
3 0
Water.

Hope it helps!
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The amount of heat absorbed when 5.5g of aluminum is heated from 25.0c to 95.0c the specific heat of aluminum is 0.897J/(gc)​
NemiM [27]

Answer:

Q=21.99 kJ

Explanation:

4 0
3 years ago
What is the hydronium ion concentration of a solution whose pH is 12.40?
pochemuha
Hydronium ion concentration = 3.98×10-13
3 0
4 years ago
Which will get spoiled faster: plum in water or plum in the open?
lara31 [8.8K]

Answer:

It depends

(plum will spoil more quickly at warm temperatures)

Explanation:

3 0
2 years ago
Phosphorus-32 has a half-life of 14.3 days. how many grams of phosphorus-32 remain after 71.5 days if you start with 4.00 grams?
almond37 [142]
Answer : 0.125 g

Explanation : If you are calculating the half life of P-32 which has half life of 14.3 days and you want to calculate how many grams are remaining after 71.3 days if yo are starting with 4 g?

starting you can calculate the number of half life P-32 undergoes first,

which is  n = 71.5 / 14.3 = 5 days.

now, to calculate the remaining mass,

m_{remaining}  = (\frac{initial mass}{2 X 2X 2X2 X2 } ) = (\frac{initial mass}{ 2^{5} } )

= = \frac{4 g }{ 2^{5} } = 0.125 g
7 0
3 years ago
The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces
Oxana [17]

Answer:

1. NaN₃(s) → Na(s) + 1.5 N₂(g)

2. 79.3g

Explanation:

<em>1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen.</em>

NaN₃(s) → Na(s) + 1.5 N₂(g)

<em>2. Suppose 43.0L of dinitrogen gas are produced by this reaction, at a temperature of 13.0°C and pressure of exactly 1atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits.</em>

First, we have to calculate the moles of N₂ from the ideal gas equation.

P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1atm.(43.0L)}{(0.08206atm.L/mol.K).286.2K} =1.83mol

The moles of NaN₃ are:

1.83molN_{2}.\frac{1molNaN_{3}}{1.5molN_{2}} =1.22molNaN_{3}

The molar mass of NaN₃ is 65.01 g/mol. The mass of NaN₃ is:

1.22mol.\frac{65.01g}{mol} =79.3g

5 0
3 years ago
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