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2 years ago

A gas can holds 2.9 gal of gasoline. What is this quantity in cubic centimeters?

Chemistry

1 answer:

DedPeter [7]2 years ago

4 0

1 gal ----------- 3785.41 cm³
2.9 gal ---------- ??

2.9 x 3785.41 / 1 => 10977.689 cm³

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What mass of carbon dioxide could be made from 100 tonnes of calcium carbonate?

MA_775_DIABLO [31]

Total mass of CaCO3 = 40 amu of Ca + 12amu of C + 16×3 amu of oxygen = 100amu of CaCO3

i.e 100 tonnes of CaCO3 .

mass of CO2 = 12amu of C + 2× 16amu of O = 44 amu of CO2

mass % of CO2 in CaCO3 = (44/100)×100 =44%

i.e
44% of 100 tonnes is CO2.
=44 tonnes of CO2.

therefore, 44% of CO2 is present in CaCO3.

3 0

1 year ago

What are the missing coefficients for the skeleton equation below? cr(s) + fe(no3)2(aq) → fe(s) + cr(no3)3(aq)?

strojnjashka [21]

SThe missing coefficient for the skeleton equation below is as follows

skeleton equation

Cr(s) + Fe(No3)2(aq) ------> Fe (s) + Cr(NO3)3 (aq)
the missing coefficient are is as follows

2 Cr(s) + 3 Fe(NO3)2 ---> 3 Fe (s) + 2 Cr(NO3)3

This is obtained by making sure all the molecules are balanced in both sides

8 0

2 years ago

If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter incr

Alinara [238K]

Answer:

The value of the heat capacity of the Calorimeter $C_c$ = 54.4 $\frac{J}{c}$

Explanation:

Given data

Heat added Q = 4.168 KJ = 4168 J

Mass of water $m_w$ = 75.40 gm

Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c

From the given condition

Q = $m_w C_w$ ΔT + $C_c$ ΔT

Put all the values in above equation we get

4168 = 75.70 × 4.18 × 11.24 + $C_c$ × 11.24

611.37 = $C_c$ × 11.24

$C_c$ = 54.4 $\frac{J}{c}$

This is the value of the heat capacity of the Calorimeter.

7 0

1 year ago

A metallic object holds a charge of −4.8 × 10−6 C. What total number of electrons does this represent? (e = 1.6 × 10−19 C is the

vagabundo [1.1K]

Answer:

$n=3.0\times 10^{13}$

Explanation:

Charge on 1 electron = $-1.6\times 10^{-19}\ C$

The expression for charge is:-

$Charge=n\times q_e$

Given that:- Charge = $-4.8\times 10^{-6}\ C$

$-4.8\times 10^{-6}=n\times (-1.6\times 10^{-19})$

$n=\frac{4.8\times 10^{-6}}{1.6\times 10^{-19}}=3.0\times 10^{13}$

Total number of electrons, n = $3.0\times 10^{13}$

5 0

1 year ago

How many moles of al2o3 can be produced from the reaction of 10.0 g of al and 19.0 of o2?

Advocard [28]

Answer:

0.185moles of Al₂O₃

Explanation:

Mass of Al = 10g

Mass of O₂ = 19g

Equation of the reaction: 4Al + 3O₂ → 2Al₂O₃

This is the balanced reaction equation.

Solution

From the given parameters, the reactant that would determine the extent of the reaction is Aluminium. It is called the limiting reagent. Oxygen is in excess and it is in an unlimited supply.

Working from the known mass to the unknown, we simply solve for the number of moles of Al using the mass given.

Then from the equation, we can relate the number of moles of Al to that of Al₂O₃ produced:

Number of moles of Al = $\frac{mass}{molar mass}$

= $\frac{10}{27}$

= 0.37mol

From the equation:

4 moles of Al produced 2 moles of Al₂O₃

0.37 mole will yield: $\frac{2 x 0.37}{4}$ = 0.185moles of Al₂O₃

8 0

1 year ago