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Colt1911 [192]
3 years ago
11

Heated bricks or blocks of iron were used long ago to warm beds. A 1.49 -kg block of iron heated to 155 Celsius would release ho

w many joules of heat as it cooled to 22 Celsius ?
Chemistry
1 answer:
AlekseyPX3 years ago
6 0
Specific heat capacity= Quantity of heat/massxΔT
Shc of iron (constant)= 0.4494J/³C for 1g
1.49kg=1490g
Q=1490x(22-155)x0.4494
Q=<span>89057.598J</span>
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What is the correct formula and charge for the polyatomic ion called acetate
DedPeter [7]

Answer:

5.03 Naming Compounds that contain Polyatomic Ions

Explanation:

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6 0
2 years ago
In order to study hydrogen halide decomposition, a researcher fills an evacuated 1.11 L flask with 0.822 mol of HI gas and allow
Elza [17]

Answer:

Kc = 168.0749

Explanation:

  •           2HI(g)     ↔    H2(g) + I2(g)

initial mol:   0.822               0          0

equil. mol: 2(0.822 - x)         x           x

∴ [ HI ]eq = 0.055 mol/L = 2(0.822 - x) / (1.11 L )

⇒ 1.644 - 2x = 0.055 * 1.11

⇒ 1.644 = 2x + 0.06105

⇒ 2x = 1.583

⇒ x = 0.7915 mol equilibrium

⇒ [ H2 ] eq = 0.7915mol / 1.11L = 0.7130 M = [ I2 ] eq

⇒ Kc = ([ H2 ] * [ I2 ]) / [ HI ]²

⇒ Kc = ( 0.7130² ) / ( 0.055² )

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4 0
3 years ago
. a large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is sub
elena-s [515]

The large piece of jewelry  that has a mass of 132.6 g and when is submerged in a graduated cylinder that initially contains 48.6 ml water and the volume increases to 61.2 ml once the piece of jewelry is submerged, has a density of: 10.523 g/ml

To solve this problem the formulas and the procedures that we have to use  are:

  • v = v(f)-v(i)
  • d = m/v

Where:

  • d= density
  • m= mass
  • v= volume
  • v(f) = final volume
  • v(i) = initial volume

Information about the problem:

  • m = 132.6 g
  • v(i) = 48.6 ml
  • v(f) = 61.2 ml
  • v = ?
  • d =?

Applying the volume formula we get:

v = v(f)-v(i)

v = 61.2 ml - 48.6 ml

v = 12.6 ml

Applying the density  formula we get:

d = m/v

d = 132.6 g/12.6 ml

d = 10.523 g/ml

<h3>What is density?</h3>

It is a physical quantity that expresses the ratio of the body mass to the volume it occupies.

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3 0
1 year ago
I NEED HELP PLEASE, THANKS! :)
marin [14]

Answer:

\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}

Explanation:

1. Calculate the moles of copper(II) hydroxide

\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}

2. Calculate the molecules of copper(II) hydroxide

\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}

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3 years ago
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Marizza181 [45]

i think its number 3

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3 years ago
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