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3 years ago

All of the stars in the sky, except for one, are so far away that they look like small points of light.

Chemistry

1 answer:

kiruha [24]3 years ago

8 0

Answer:

The sun.

Explanation:

It is closer to Earth, therefore bigger to us, and not just a small point of light.

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Calculate the empirical formula for a compound containing 67.6% Hg 10.8% S 21.6% O

pashok25 [27]

Answer:

HgSO₄

Explanation:

% => g => moles => ratio => reduce => empirical ratio

%Hg = 67.6% => 67.6g/201g/mol = 0.34mol

%S = 10.8% => 10.8g/32g/mol = 0.34mol

%O = 21.6% => 21.6g/16g/mol = 1.35mol

Hg:S:O => 0.34:0.34:1.35

Reduce to whole number ratio by dividing by the smaller mole value...

Hg:S:O => 0.34/.34:0.34/.34:1.35/.34 => Empirical Ratio = 1:1:4

∴ Empirical Formula is HgSO₄

3 0

3 years ago

1) You are asked to make 10 miles of iron (Fe) from iron oxide (Fe2O3) and excess carbon monoxide (CO). Fe2O3(s) + 3CO(g)—> 2

Inessa05 [86]

For both of them, used the balanced equation and it’s mole ratio to convert whatever you need to into moles. See the attacked work.

1) D 5 mols

2) A 0.55 mols

8 0

3 years ago

The equilibrium constant for the formation of ammonia from nitrogen and hydrogen is 1.6 × 102. what is the form of the equilibri

Nimfa-mama [501]

Answer: The expression for equilibrium constant is $\frac{[NH_3]^2}{[H_2]^3[N_2]}$

Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as $k_c$

For a general reaction:

$aA+bB\rightleftharpoons cC+dD$

The equilibrium constant is written as:

$k_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

Chemical reaction for the formation of ammonia is:

$N_2+3H_3\rightleftharpoons 2NH_3$

$k_c=1.6\times 10^2$

Expression for $k_c$ is:

$k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}$

$1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}$

8 0

3 years ago

Two solutions, initially at 24.60 °C, are mixed in a coffee cup calorimeter (Ccal = 15.5 J/°C). When a 100.0 mL volume of 0.100

yulyashka [42]

Answer:

ΔH = -59.6kJ/mol

Explanation:

The reaction that occurs between Ag⁺ and Cl⁻ ions is:

Ag⁺ + Cl⁻ → AgCl(s) + ΔH

To find ΔH we need to obtain moles of reaction and heat released in the reaction because ΔH is defined as heat released per mole of reaction.

<em>Moles of reaction:</em>

Moles of Ag⁺ and Cl⁻ added are:

Ag⁺: 0.100L * (0.100mol / L) = 0.01moles

Cl⁻: 0.100L * (0.200mol / L) 0 0.02 moles

That means limiting reactant is Ag⁺ and moles of reaction are 0.01 moles

<em>Heat released:</em>

To find heat released we must use coffe cup calorimeter equation:

Q = C*m*ΔT

<em>Where C is specific heat of solution (4.18J/g°C), m is the mass of solution (200g because there are 100 + 100mL = 200mL and density of solution is 1g/mL) and ΔT is change in temperature (25.30°C - 24.60°C = 0.70°C).</em>

Replacing:

Q = C*m*ΔT

Q = 4.18J/g°C * 200g * 0.70°C

Q = 585,2J

Is total heat released.

The calorimeter absorbs:

15.5J / °C * 0.7°C = 10.85

Thus, when 0.01 moles reacts, 585.2J + 10.85 = 596.05J are released (Heat released is heat abosrbed by calorimeter + Heat absorbed by water) and ΔH is:

ΔH = 596.05J / 0.01 moles =

ΔH = 59605J / mol =

<em>As heat is released, ΔH < 0.</em>

6 0

3 years ago

What data should be plotted to show that experimental concentration data fits a first-order reaction? A) 1/[reactant] vs. time B

natita [175]

Answer:

C) In[reactant] vs. time

Explanation:

For a first order reaction the integrated rate law equation is:

$A = A_{0}e^{-kt}$

where A(0) = initial concentration of the reactant

A = concentration after time 't'

k = rate constant

Taking ln on both sides gives:

$ln[A] = ln[A]_{0}-kt$

Therefore a plot of ln[A] vs t should give a straight line with a slope = -k

Hence, ln[reactant] vs time should be plotted for a first order reaction.

7 0

3 years ago