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serg [7]
3 years ago
13

The safety air bags in automobiles are inflated by nitrogen gas generated by rapid decomposition of sodium azide (NaN 3).

Chemistry
1 answer:
schepotkina [342]3 years ago
5 0
Or, you can have no airbags and die... hehe
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Which statements about salt water are true? Check all that apply.
Ymorist [56]

Salt water contains minerals other than salt.

The temperature of ocean water is affected by the depth of the water.

The temperature of ocean water is affected by the ocean’s location.

Objects float more easily in oceans than in lakes or rivers.

These are all of the true answers

6 0
3 years ago
Read 2 more answers
A sealed helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 293 K is moved. What volume will the bal
bixtya [17]

Answer:

6.48 L

Explanation:

From the question,

Applying

PV/T = P'V'/T'......................... Equation 1

P = initial pressure of the helium balloon, V =  Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.

make V' the subject of the equation

V' = PVT'/P'T......................... Equation 2

Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm

Substitute these values into equation 2

V' = (4.5×1×253)/(0.6×293)

V' = 1138.5/175.8

V' = 6.48 L

8 0
3 years ago
8. Bring the balloon in contact with the wall. What happens to the charges in the wall?
Ne4ueva [31]

When the charged balloon is brought near the wall, it repels some of the negatively charged electrons in that part of the wall. Therefore, that part of the wall is left repelled.

<u>Explanation</u>:  

  • Balloons don't stick to walls. However, if you rub the balloon on an appropriate piece of material such as clothing or a wall, electrons are pulled from the other material to the balloon.
  • The balloon now as more electrons than normal and therefore has an overall negative charge. Two balloons like this will repel each other.
  • The other material now has an overall positive charge. Because opposite charges attract, the balloon will now appear to stick to the other material. If you didn't rub the balloon first, it's charge would be neutral and it wouldn't stick to the wall.
7 0
3 years ago
The combustion of 135 mg of a hydrocarbon produces 440 mg of CO2 and 135 mg H2O. The molar mass of the hydrocarbon is 270 g/mol.
NeX [460]

Answer:

Molecular formula = C20H30

Explanation:

NB 440mg = 0.44g, 135mg= 0.135g

From the question, moles of CO2= 0.44/44= 0.01mol

Since 1 mol of CO2 contains 1mol of C, it implies mol of C = 0.01

Also from the question, moles of H2O = 0.135/18= 0.0075mole

Since 1 mol of H2O contains 2mol of H, it implies mol of H = 0.0075×2= 0.015 mol of H

To get the empirical formula, divide by smallest number of mole

Mol of C = 0.01/0.01=1

Mol of H = 0.015/0.01= 1.5

Multiply both by 2 to obtain a whole number

Mol of C =1×2 = 2

Mol of H= 1.5×2 = 3

Empirical formula= C2H3

[C2H3] not = 270

[ (2×12) + 3]n = 270

27n = 270

n=10

Molecular formula= [C2H3]10= C20H30

5 0
3 years ago
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°
miskamm [114]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1835 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)    \Delta H_1=157kJ   ( × 4)

(2) P_4(g)+6Cl_2(g)\rightarrow 4PCl_3(g)     \Delta H_2=-1207kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(4\times (-157))+(1\times (-1207))=-1835kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1835 kJ.

6 0
3 years ago
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