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jeka57 [31]
3 years ago
11

Which two columns have identical data? What do these two columns suggest about the identity of elements?

Chemistry
1 answer:
Gnesinka [82]3 years ago
3 0

Answer:

Explanation:

Atomic number and number of protons have identical data. These two columns suggest that the atomic number is the same as the number of protons in an atom.

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AgI + KNO3<br>What type of reaction is this?
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Answer:

double replacement

Explanation:

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6. A piece of solid gold was heated from 274K to 314K. 35.73 of energy was needed to raise the temperature.
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From Q = mcΔT, we can rearrange the equation to solve for mass, m = Q/cΔT. The specific heat capacity, c, of solid gold is 0.129 J/g °C. I'm assuming that the energy is given in joules, as it's not specified in the question as written.

m = Q/cΔT = (35.73 J)/(0.129 J/g °C)(40.85 °C - 0.85°C)

m = 6.92 g of gold was present  

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What is the formula of a compound formed between nitrogen (N) and barium (Ba)?
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A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions
Mandarinka [93]

Answer:

\boxed{\text{-55.8 kJ/mol NaOH}}

Explanation:

NaOH + HNO₃ ⟶ NaNO₃ + H₂O

There are two energy flows in this reaction.

\begin{array}{cccl}\text{Heat from neutralization} & + &\text{Heat absorbed by water} & = 0\\q_{1} & + & q_{2} & =0\\n\DeltaH & + & mC\Delta T & =0\\\end{array}

Data:

V(base) = 100.0 mL; c(base) = 0.300 mol·L⁻¹

V(acid) = 100.0 mL; c (acid) = 0.300 mol·L⁻¹  

        T₁ = 35.00 °C;          T₂ = 37.00 °C

Calculations:

(a) q₁

n_{\text{NaOH}} = \text{0.1000 L } \times \dfrac{\text{0.300 mol}}{\text{1 L}} = \text{0.0300 mol}\\\\n_{\text{HNO}_{3}} = \text{0.1000 L } \times \dfrac{\text{0.300 mol}}{\text{1 L}} = \text{0.0300 mol}

We have equimolar amounts of NaOH and HNO₃

n = 0.0300 mol

q₁ = 0.0300ΔH

(b) q₂

 V = 100.0 mL + 100.0 mL = 200.0 mL

m = 200.0 g

ΔT = T₂ - T₁ = 37.00 °C – 35.00 °C = 2.00 °C

q₂ = 200.0 × 4.184 × 2.00 = 1674 J

(c) ΔH

0.0300ΔH + 1674 = 0

          0.0300ΔH = -1674

                      ΔH = -1674/0.0300

                      ΔH = -55 800 J/mol

                      ΔH = -55.8 kJ/mol

\Delta_{r}H^{\circ} = \boxed{\textbf{-55.8 kJ/mol NaOH}}

6 0
3 years ago
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