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3 years ago

How many significant figures are there in the measurement 0.003kg

Chemistry

1 answer:

salantis [7]3 years ago

6 0

Answer:

-4

Explanation:

In 0.0034 the zeros are preceding numberd and after the decimal point.

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When iron reacts with oxygen, it forms iron oxide, or rust.

Feliz [49]

Answer:

Mass of iron oxide: 79.85 g

Explanation:

Given data:

Mass of iron = 112 g

Mass of O₂ = 24 g

Mass of iron oxide formed = ?

Solution:

4Fe + 3O₂ → 2Fe₂O₃

Number of moles of Fe:

Number of moles = mass/ molar mass

Number of moles = 112 g/ 55.85 g/mol

Number of moles = 2 mol

Number of moles of O₂:

Number of moles = mass/ molar mass

Number of moles = 24 g/ 32 g/mol

Number of moles = 0.75 mol

Now we will compare the moles of oxygen and iron with iron oxide.

Fe : Fe₂O₃

4 : 2

2 : 2/4×2 = 1

O₂ : Fe₂O₃

3 : 2

0.75 : 2/3×0.75 = 0.5

The number of moles of iron oxide formed by oxygen are less thus it will be limiting reactant.

Mass of iron oxide:

Mass = number of moles × molar mass

Mass = 0.5 mol × 159.69 g/mol

Mass = 79.85 g

3 0

3 years ago

Oxidation of Nitrogen in N2O3

kirill [66]

Explanation:

<h3>oxidation of Nitrogen in N2O3 is </h3><h2>+3</h2>

7 0

2 years ago

Suppose that you take 200 mg of an antibiotic every 8 hr. The half-life of the drug is 8 hr (the time it takes for half of the

DochEvi [55]

Answer:

600 mg

Explanation:

The initial amount of the drug = 200 mg

The half-life of the drug = 8 hrs

It means that:-

After 6 hours, the concentration becomes :- $\frac{200}{2}$ mg

After 12 hours, the concentration becomes :- $\frac{200}{4}$ mg

After 18 hours, the concentration becomes :- $\frac{200}{8}$ mg

And so on...

Thus,

After infinite time = $200+\frac{200}{2}+\frac{200}{4}+\frac{200}{8}+..$

Thus,

After infinite time = $200\times (1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..)$

The sum of the infinite series is:- $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..$ = $\frac{1}{1+\frac{1}{2}}=2$

So,

<u>After infinite time = $200\times 2$ mg = 600 mg</u>

5 0

3 years ago

At STP, what volume of container would be needed to hold 3.2 moles of gas?

Rzqust [24]

78.4 L volume of container is required to hold 3.2 moles of gas.

Explanation:

STP is defined as the standard temperature and pressure of a gas in room temperature conditions. At STP, one mole of the gas which has Avogadro's number of molecules in it will occupy a volume of 22.4 L.
So, one mole of a substance or gas will occupy a volume of 22.4 L then the volume of the container needed for 3.2 moles of gas is calculated by multiplying 22.4 L, standard volume with the moles of the gas 3.2 moles.
Hence, the answer would be 78.4 L.

3 0

3 years ago

At 311 K, this reaction has a K c value of 0.0111 . X ( g ) + 2 Y ( g ) − ⇀ ↽ − 2 Z ( g ) Calculate K p at 311 K. Note that the

Aleks [24]

Answer:

$K_{p}=4.35\times 10^{-4}$

Explanation:

We know, $K_{p}=K_{c}(RT)^{\Delta n}$

where, R = 0.0821 L.atm/(mol.K), T is temperature in kelvin and $\Delta n$ is difference in sum of stoichiometric coefficient of products and reactants

Here $\Delta n=(2)-(2+1)=-1$ and T = 311 K

So, $K_{p}=(0.0111)\times [(0.0821L.atm.mol^{-1}.K^{-1})\times 311K]^{-1}=4.35\times 10^{-4}$

Hence value of equilibrium constant in terms of partial pressure $(K_{p})$ is $4.35\times 10^{-4}$

4 0

3 years ago