All categories

3 years ago

When making calculations, you should rely on the precision of your measured data.

Chemistry

1 answer:

ELEN [110]3 years ago

3 0

True

Im happy i could help you today

have a great rest of ur week :)

You might be interested in

In a certain compound of iron and oxygen, FexOy, it is found that a sample of this compound weighing 6.285 g contains 4.396 g of

saveliy_v [14]

Answer:

Formula of oxide is $Fe_{2}O_{3}$

Explanation:

The given compound consists of Fe and O.

So, mass of oxygen in sample = (mass of sample) - (mass of Fe in sample)

= (6.285 g ) - (4.396 g)

= 1.889 g

Molar mass of O = 16 g/mol and molar mass of Fe = 55.845 g/mol

So, ratio of number of moles of Fe and O (Fe : O)

= $\frac{4.396}{55.845}:\frac{1.889}{16}$

= 0.0787 : 0.118

= $1:\frac{0.118}{0.0787}$

= 1 : 1.5

= 2 : 3

So, formula of oxide is $Fe_{2}O_{3}$

5 0

3 years ago

Provided is a diagram showing forces on a box.

Eduardwww [97]

Answer:

5 N left

Explanation:

trust me cccuuhh

4 0

3 years ago

Write the name of the compounds

blagie [28]

P2O5 = Phosphorus pentoxide

CuO = Copper (II) oxide

NH4CI = Ammonium Chloride

Mn(OH)2 = Pyrochroite

H2O2 = Hydrogen peroxide

P4S9 = Tetraphosphorus nonasulfide

CIO2 = Chlorine dioxide

NaF = Sodium fluoride

FeSO3 = Iron (II) Sulfite

Fe(NO3)3 = Iron (III) Nitrate

Cr(NO2)3 = Chromium (III) Nitrite

NaHCO3 = Sodium Hydrogen Carbonate

H2PO4 = Dihydrogen Phosphate Ion

NaCN = Sodium Cyanide

IF7 = Iodine Heptafluoride

PCI3 = Phosphorus Trichloride

5 0

3 years ago

If you perfom 30 joules of work lifting a 20-N box from the floor to a shelf how high is the shelf

Ede4ka [16]

Answer: 1.5 m

Explanation: E=mgh=Fh

30 J = 20 N * h

h = 30/20 = 1.5 m

5 0

3 years ago

The standard molar enthalpy of formation of NH3(g) is -45.9 kJ/mol. What is the enthalpy change if 9.51 g N2(g) and 1.96 g H2(g)

Vladimir [108]

Answer:

$\Delta H=-29.7kJ$

Explanation:

Hello!

In this case, since the undergoing chemical reaction is:

$N_2+3H_2\rightarrow 2NH_3$

We first need to identify the limiting reactant given the masses of nitrogen and hydrogen:

$n_{NH_3}^{by\ H_2}=1.96gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.647molNH_3\\\\ n_{NH_3}^{by\ N_2}=9.51gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.679molNH_3$

It means that only 0.647 moles of ammonia are yielded, so the resulting enthalpy change is:

$\Delta H=0.647molNH_3*\frac{-45.9kJ}{1molNH_3}\\\\ \Delta H=-29.7kJ$

Best regards!

3 0

3 years ago