Answer:
0.0761.
0.0059.
Explanation:
Hello!
In this case, since the writing of numbers in standard notation given the scientific notation requires adding zeros to the left of the first nonzero digit, for the first number, we need to add two zeros as seven is the third digit position:
0.0761.
Next, we need to add three zeros as 5 would be the fourth digit:
0.0059.
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22.3 g of NaN₃ are required to fully inflate an airbag of 11.6 L at STP.
To find the mass, the given data was,
Volume = 11.6 Liters
<h3>What is decomposition reaction?</h3>
A decomposition reaction can be defined as a chemical reaction in which one reactant breaks down into two or more products.
In airbags, sodium azide decomposes to form sodium and nitrogen gas, which inflates the bag. The decomposition reaction is:
2 NaN₃ ⇒ 2 Na + 3 N₂
We can calculate the mass of NaN₃ needed to produce 11.6 L of N₂ at STP, using the following relations.
- At STP, 1 mole of N₂ occupies 22.4 L.
- The molar ratio of N₂ to NaN₃ is 3:2.
- The molar mass of NaN₃ is 65.01 g/mol.
Substituting all the known values to find the volume,
11.6 × ( 1 / 22.4) × ( 2/3) × ( 65.01 / 1)
= 22.4 g.
22.4 g of NaN₃ are required to fully inflate an airbag of 11.6 L at STP.
Learn more about decomposition reaction,
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Answer:
Explanation:
pH = pKa + log [ CH₃COO⁻ ] / [CH₃COOH ]
5.36 = 4.86 + log [ CH₃COO⁻ ] / [CH₃COOH ]
log [ CH₃COO⁻ ] / [CH₃COOH ] = .5
[ CH₃COO⁻ ] / [CH₃COOH ] = 3.16
moles of CH₃COOH = .680 x .9 = .612 M
Let x mole of KOH is required
x /( .612 - x ) = 3.16
x = 1.933 - 3.16 x
x = .46488
.46488 moles of KOH will be required
volume required be V
v x 2.62 = .46488
v = .1774 L
= 177.4 mL
177.4 mL of 2.62 M KOH will be required .
Answer:
Explanation:
1. Calculate the rate constant
The integrated rate law for first order decay is
where
A₀ and A_t are the amounts at t = 0 and t
k is the rate constant
2. Calculate the half-life
They contain the same number of molecules; however, the masses can differ depending on the molar masses of the substances.
