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3 years ago

Give an example of at least one element made up of molecules and one compound made up of molecules.

Chemistry

1 answer:

zavuch27 [327]3 years ago

8 0

I think mercury for element and sodium hypochlorite

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At a certain temperature and pressure, one liter of CO2 gas weighs 1.95 g.

AysviL [449]

Answer:

1.332 g.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

At the same T and P and constant V (1.0 L), different gases have the same no. of moles (n):

∴ (n) of CO₂ = (n) of C₂H₆

∵ n = mass/molar mass

∴ (mass/molar mass) of CO₂ = (mass/molar mass) of C₂H₆

mass of CO₂ = 1.95 g, molar mass of CO₂ = 44.01 g/mol.

mass of C₂H₆ = ??? g, molar mass of C₂H₆ = 30.07 g/mol.

∴ mass of C₂H₆ = [(mass/molar mass) of CO₂]*(molar mass) of C₂H₆ = [(1.95 g / 44.01 g/mol)] * (30.07 g/mol) = 1.332 g.

7 0

3 years ago

Find the percent composition of OXYGEN in Manganese (III) nitrate, Mn(NO3)3.

BaLLatris [955]

Answer:

59.8%

Explanation:

First find the Mr of manganese (III) nitrate.

Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = 240.9

Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:

Mr of (O₃)₃ = (16 × 3) × 3 = 144

Now we can find percentage composition / percentage by mass of oxygen.

% composition = $\frac{Mr\ of\ oxygen\ in\ compound}{Mr\ of\ compound}$ × 100

% composition = $\frac{144}{240.9}$ × 100 = 59.776%

∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).

8 0

2 years ago

The decomposition of N_2O_5(g) following 1st order kinetics. N_2O_5(g) to N_2O_4(g) + ½ O_2(g) If 2.56 mg of N_2O_5 is initially

Crank

Answer: The rate constant is $0.334s^{-1}$

Explanation ;

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = ?

t = age of sample = 4.26 min

a = initial amount of the reactant = 2.56 mg

a - x = amount left after decay process = 2.50 mg

Now put all the given values in above equation to calculate the rate constant ,we get

$4.26=\frac{2.303}{k}\log\frac{2.56}{2.50}$

$k=\frac{2.303}{4.26}\log\frac{2.56}{2.50}$

$k=5.57\times 10^{-3}min^{-1}=5.57\times 10^{-3}\times 60s^{-1}=0.334s^{-1}$

Thus rate constant is [tex]0.334s^{-1}

4 0

3 years ago

Oxidation of Nitrogen in NO2

tiny-mole [99]

Explanation:

oxidation of Nitrogen in NO2 is +4

8 0

2 years ago

How do electrons behave?

agasfer [191]

An electron in motion generates an electromagnetic field and is in turn deflected by external electromagnetic fields. When an electron is accelerated, it can absorb or radiate energy in the form of photons. Electrons, together with atomic nuclei made up of protons and neutrons, make up the

5 0

3 years ago