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3 years ago

6

Sunlight shining through a clear window it’s Jeremy’s face while he does his homework he uses a large wooden wooden block to cov

er the window what do the wooden block in the clear window demonstrate

2 answers:

lidiya [134]3 years ago

4 0

Answer:

a blocker?

Explanation:

just a guess

Deffense [45]3 years ago

3 0

I would have to know the subject

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How many grams are in 10.25 moles of zinc chromate?

Kobotan [32]

Answer:

1859.4 g of ZnCrO₄ in 10.25 moles

Explanation:

First of all, we determine the molecular formula of the compound:

Zinc → Zn²⁺ (cation)

Chromate → CrO₄⁻² (anion)

Zinc chromate → ZnCrO₄

Molar mass for the compound is:

Molar mass of Zn + Molar mass of Cr + (Molar mass of O) . 4 = 181.41 g/mol

65.41 g/mol + 52 g/mol + 16 g/mol . 4 = 181.41 g/mol

Let's apply this conversion factor: 10.25 mol . 181.41 g/mol = 1859.4 g

5 0

3 years ago

A helium gas balloon is expanded to 78.0 L, while the pressure is held constant at 0.37 atm. If the work done on the gas mixture

Elenna [48]

Answer:

77.248 L

Explanation:

From the question,

Work done on the gas mixture is given as,

W = PΔV.................. Equation 1

Where W = work done, P = pressure of the the gas, ΔV = Change in volume of the gas.

make ΔV the subject of the equation

ΔV = W/P..................... Equation 2

Given: W = 28.2 J, P = 0.37 atm = (0.37×101325) N/m² = 37490.25 N/m²

Substitute into equation 2

ΔV = 28.2/37490.25

ΔV = 0.000752 m³

ΔV = 0.752 L

But,

ΔV = V₂-V₁................. Equation 3

Where V₂ = Final volume of the helium gas, V₁ = Initial volume of the helium gas

make V₁ the subject of the equation

V₁ = V₂-ΔV................ Equation 4

Given: V₂ = 78 L.

Substitute into equation 4

V₁ = 78-0.752

V₁ = 77.248 L

8 0

3 years ago

Read 2 more answers

How many chlorine atoms would be in 6.02 X 10^23 units of gold III chloride

Pepsi [2]

Answer:

The number of chlorine atoms present in $6.02 \times 10^{23}$ units of gold III chloride is $18.066 \times 10^{23}$

Explanation:

Formula of Gold (III) chloride: $AuCl_{3}$

<em>Avogadro Number</em> : Number of particles present in <u>one mole</u><u> </u>of a substance.

${N_{0}} =6.022 \times 10^{23}$

Using,

$n(moles)=\frac{Given\ number\ of\ particles}{N_{0}}$

$n =\frac{6.02\times 10^{23}}{6.022\times 10^{23}}$

= 1 mole(0.9999 , nearly equal to 1 )

The given Gold III chloride sample is 1 mole in amount.

$6.022 \times 10^{23}$ = 1 mole of $AuCl_{3}$

In this Sample,

1 mole of $AuCl_{3}$ will give = 3 mole of Chlorine atoms

1 mole of Cl contain = $6.022 \times 10^{23}$

3 mole of Cl contain = $6.022 \times 10^{23}\times 3$

3 mole of Cl contain = $18.066 \times 10^{23}$

So,

The number of chlorine atoms present in $6.02 \times 10^{23}$ units of gold III chloride is $18.066 \times 10^{23}$

8 0

3 years ago

A 10.000g ice cube is added to 100.000g of water at 80.0 oC in a calorimeter. The final temperature of the water in the calorime

Ket [755]

<u>Answer:</u> The amount of heat needed to melt the ice cube is 6.083 kJ

<u>Explanation:</u>

The processes involved in the given problem are:

$1.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\2.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(65.5^oC,338.5K)$

<u>For process 1:</u>

To calculate the amount of heat required to melt the ice at its melting point, we use the equation:

$q_1=m\times \Delta H_{fusion}$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of ice = 10.000 g

$\Delta H_{fusion}$ = enthalpy change for fusion = 334.16 J/g

Putting all the values in above equation, we get:

$q_1=10.000g\times 334.16J/g=3341.6J$

<u>For process 2:</u>

To calculate the heat required at different temperature, we use the equation:

$q_2=mc\Delta T$

where,

$q_2$ = heat absorbed

m = mass of ice = 10.000 g

c = specific heat capacity of water = 4.184 J/g.°C

$\Delta T$ = change in temperature = $T_2-T_1=[65.0-0]^oC=65.5^oC$

Putting values in above equation, we get:

$q_2=10.000g\times 4.184J/g.^oC\times 65.5^oC\\\\q_2=2741.83J$

Total heat absorbed = $q_1+q_2$

Total heat absorbed = $[3341.6+2741.83]J=6083.43J=6.083kJ$

Hence, the amount of heat needed to melt the ice cube is 6.083 kJ

4 0

3 years ago

Which of the following statements is not true about the decomposition of a simple binary compound?

Brilliant_brown [7]

The products are unpredictable is your answer .-/

6 0

3 years ago

Read 2 more answers