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earnstyle [38]
2 years ago
3

The material typically used to heat metal radiators is water. J of heat must be supplied to the radiator in order to raise its t

emperature from 25.0 to 55.0°C. If a boiler generates water at 84.0°C, what mass of water is required to heat the radiator? Water has a specific heat of 4.184 .
Chemistry
1 answer:
LekaFEV [45]2 years ago
8 0

Complete question:

The material typically used to heat metal radiators is water. 4.05 x 10⁵ J of heat must be supplied to the radiator in order to raise its temperature from 25.0 to 55.0°C. If a boiler generates water at 84.0°C, what mass of water is required to heat the radiator? Water has a specific heat of 4.184 J.g⁻¹.C⁻¹.

Answer:

The mass of water required to heat the radiator is 3.338 kg

Explanation:

Given;

Quantity of heat supplied to raise the temperature of radiator, Q = 4.05 x 10⁵ J

Specific heat capacity of water, c = 4.184 J.g⁻¹.C⁻¹ = 4.184 x 10³ J.kg⁻¹C⁻¹

Quantity of heat supplied to raise the temperature of water;

Q = mcΔθ

where;

m is the mass of water

c is the specific heat capacity of water

Δθ is the temperature difference of water from 84.0°C to 55.0°C

m = Q / cΔθ ,        Δθ = 84.0°C - 55.0°C = 29°C

m = (4.05 x 10⁵ ) / (4.184 x 10³ x 29)

m = 3.338 kg

Therefore, the mass of water required to heat the radiator is 3.338 kg

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3 0
2 years ago
The interior of a refrigerator has a volume of 0.600 m3. the temperature inside the refrigerator in 282 k, and the pressure is 1
Dafna11 [192]
In this instance we can use the ideal gas law equation to find the number of moles of gas inside the refrigerator 
PV = nRT
where 
P - pressure - 101 000 Pa
V - volume - 0.600 m³
n - number of moles
R - universal gas constant - 8.314 J/mol.K
T - temperature - 282 K
substituting these values in the equation 
101 000 Pa x 0.600 m³ = n x  8.314 J/mol.K x 282 K
n = 25.8 mol
there are 25.8 mol of the gas
to find the mass of gas
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6 0
2 years ago
The empirical formula of a gaseous fluorocarbon is CF2. At a certain temperature and pressure, a 1-L volume holds 8.93 g of this
dimaraw [331]

Answer:

C₄F₈

Explanation:

Using their mole ratio to compute their mass

molar mass of carbon = 12.0107 g/mol

molar mass of fluorine gas = 37.99681

let x = mass of carbon

given mass of fluorine = 1.70 g

x / 12.01067 = 1.70 / 37.99687

cross multiply

x = ( 1.70 × 12) / 37.99687 = 20.4 / 37.99687 = 0.53688 g

mass of one mole of CF₂ = 0.53688 + 1.70 = 2.23688 g

number of mole of CF₂ = 8.93 g / 2.23688 = 3.992 approx 4

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3 0
2 years ago
The piece of iron that miguel measured had a mass of 51.1 g and a volume of 6.63 cm 3 . what did miguel calculate to be the dens
likoan [24]
Given:
Mass, m = 51.1 g
Volume, V = 6.63 cm³

By definition, 
Density = Mass/Volume
              = (51.1 g)/(6.63 cm³)
              = 7.7074 g/cm³

In SI units,
Density = (7.7074 g/cm³)*(10⁻³ kg/g)*(10² cm/m)³
              = 7707.4 kg/m³

Answer: 7.707 g/cm³ or 7707.4 kg/m³

4 0
2 years ago
What is the net ionic equation of the reaction of MgCl2 with NaOH? Express your answer as a chemical equation. View Available Hi
kari74 [83]

Answer:

Net ionic equation for the reaction between MgCl₂ and NaOH in water:

\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s).

Net ionic equation for the reaction between MgSO₄ and BaCl₂ in water:

\rm {Ba}^{2+}\; (aq) + {SO_4}^{2-}\;(aq) \to BaSO_4\; (s).

Explanation:

Start by finding the chemical equations for each reaction:

MgCl₂ reacts with NaOH to form Mg(OH)₂ and NaCl. This reaction is a double decomposition reaction (a.k.a. double replacement reaction, salt metathesis reaction.) This reaction is feasible because one of the products, Mg(OH)₂, is weakly soluble in water and exists as a solid precipitate.

\rm MgCl_2\; (aq) + 2\; NaOH\; (aq)\to Mg(OH)_2 \; (s) + 2\; NaCl\; (aq).

MgSO₄ reacts with BaCl₂ in a double decomposition reaction to produce BaSO₄ and MgCl₂. Similarly, the solid product BaSO₄ makes this reaction is feasible.

\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq).

How to rewrite a chemical equation to produce a net ionic equation?

  1. Rewrite all reactants and products that ionizes completely in the solution as ions.
  2. Eliminate ions that exist on both sides of the equation to produce a net ionic equation.

Typical classes of chemicals that ionize completely in water:

  • Soluble salts,
  • Strong acids, and
  • Strong bases.

Keep the formula of salts that are not soluble in water, weak acids, weak bases, and water unchanged.

Take the first reaction as an example, note the coefficients:

  • MgCl₂ is a salt and is soluble in water. Each unit of MgCl₂ can be written as \rm Mg^{2+} and \rm 2\; Cl^{-}.
  • NaOH is a strong base. Each unit of NaOH can be written as \rm Na^{+} and \rm OH^{-}.
  • Mg(OH)₂ is a weak base and should not be written.
  • NaCl is a salt and is soluble in water. Each unit of NaCl can be written as \rm Na^{+} and \rm Cl^{-}.

\rm Mg^{2+} + 2\; Cl^{-} + 2\; Na^{+} + 2\; OH^{-} \to Mg(OH)_2\;(s) + 2\; Na^{+} + 2\; Cl^{-}.

Ions on both sides of the equation:

  • \rm 2\; Cl^{-}, and
  • \rm 2\; Na^{+}.

Add the state symbols:

\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s).

For the second reaction:

\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq).

\rm Mg^{2+} + 2\; {SO_4}^{2-} + Ba^{2+} + 2\; Cl^{-} \to BaSO_4\; (s) + Mg^{2+} + 2\; Cl^{-}.

\rm Ba^{2+}\; (aq) + {SO_4}^{2-}\; (aq) \to BaSO_4\; (s).

5 0
2 years ago
Read 2 more answers
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