Answer:
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Answer:
2 significant figures i.e. 2 and 7.
Explanation:
We need to find the number of significant figures in the given number i.e.
.
It has 2 significant figures i.e. 2 and 7.
It means the power of 10 doesn't count in the significant figures.
Hence, there are 2 significant figures.
Answer:
The percentage by mass of benzene in the solution is approximately 0.2%
Explanation:
The given parameters are;
The mass of the benzene solute dissolved in the gasoline solvent, m₁ = 1.56 g
The total volume of the benzene gasoline solution made, V = 998.44 mL
The density of gasoline, ρ = 0.7489 g/mL
Mass, m = Density, ρ × Volume, V
∴ The mass of gasoline in the 998.44 mL, solution = 0.7489 g/mL × 998.44 mL = 747.731716 g
The total mass of the solution = The mass of the benzene in the solution + The mass of the gasoline in the solution
∴ The total mass of the solution = 747.731716 g + 1.5 g = 749.231716 g

The percentage by mass of benzene in the solution = (1.5 g/749.231716 g)×100 ≈ 0.2% by mass.
Gamma rays can transmit through matter.
Answer:
At equilibrium:
[H2] = 0.005 M
[Br2] = 0.105 M
[HBr] = 0.189 M
Explanation:
H2(g) + Br2(g) ⇄ 2HBr
an "x" value will be used from reactant to produced "2x"
so at equilibrium:
[H2] = 0.1 - x
[Br2] = 0.2 - x
[HBr] = 2x
we know that Kc=[HBr]²/[H2][Br2]
Thus 62.5 = (2x)²/(0.1-x)(0.2-x)
this generate a quadratic equation: 58.5x² - 18.75x + 1.25 = 0
the x₁ = 0.23 x₂ = 0.09457
we pick 0.09457 because the two reactants can not make more than what they have. x₁ is higher than both initial reactant concentration
Then we substitute the "x₂" value at equilibrium:
[H2] = 0.1-0.09457 = 0.005 M
[Br2] = 0.2-0.09457 = 0.105 M
[HBr] = 2*0.09457 = 0.189 M