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Tema [17]
3 years ago
6

Round to 4 significant figures. 0.007062

Chemistry
1 answer:
Monica [59]3 years ago
7 0
.0071 you round 0 to1 because the 6 is bigger than 5 and if anything is bigger than five you round up that last number
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adaptation may be i guess

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2 years ago
What is the initial temperature of 50g of water that was raised to 50*C by the addition of 4.18kJ of heat energy? (2 sig fig)
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Answer:

50

Explanation:

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3 years ago
Upon balancing the equation below, how many moles of sulfuric acid are needed to react completely with 8.4 moles of lithium hydr
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The answer is b. 4.2 mole. The balanced reaction formula is 2LiOH + H2SO4 -->Li2SO4 + 2H2O. And the ratio of mole number of the reactants is the same as the ratio of coefficients.
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3 years ago
8. Calculate the number of moles of eachsubstance.a. 5.45 x 1026 particles of methane, CH4
Klio2033 [76]

<em>ANSWER</em>

The number of moles of methane is 905.32 moles

STEP-BY-STEP EXPLANATION:

Given information

The number of particles of methane = 5.45 x 10^26 particles

Let x represents the number of moles of methane

To calculate the number of moles, we will be using the below formula

\text{Number of particles = number of moles x Avogadro's constant}

Recall that, the Avogadro's constant is given as

6.02\cdot10^{23}\begin{gathered} 5.45\cdot10^{26}\text{ = x }\cdot\text{ 6.02 }\cdot10^{23} \\ \text{Divide both sides by 6.02 }\cdot10^{23} \\ x\text{ = }\frac{5.45\cdot10^{26}}{6.02\cdot10^{23}} \\ x\text{ = }\frac{5.45}{6.02}\cdot10^{26\text{ - 23}} \\ x\text{ = 0.9053 }\cdot10^3 \\ x\text{ = 905.32 moles} \end{gathered}

Therefore, the number of moles of methane is 905.32 moles

6 0
1 year ago
NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.084 M in
kolbaska11 [484]
Answer is: pH of solution is 5,17.
Kb(NH₃) = 1,8·10⁻⁵.
c(NH₄Cl) = 0,084 M = 0,084 mol/L.
Chemical reaction: NH₄⁺ + H₂O → NH₃ + H₃O⁺.
Ka · Kb = 10⁻¹⁴.
Ka(NH₄⁺) = 10⁻¹⁴ ÷ 1,8·10⁻⁵.
Ka(NH₄⁺) = 5,55·10⁻¹⁰.
[H₃O⁺] = [NH₃]  = x.
Ka(NH₄⁺) = [H₃O⁺] · [NH₃] ÷ [NH₄⁺].
5,55·10⁻¹⁰ = x² ÷ (0,084 M - x).
Solve quadratic equation: x = [H₃O⁺] = 6,8·10⁻⁶ M.
pH = -log[H₃O⁺].
pH = -log(6,8·10⁻⁶ M) = 5,17.
7 0
3 years ago
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