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3 years ago

Leslie incorrectly balances an equation as 2C4H10 + 12O2 → 8CO2 + 10H2O.

Chemistry

2 answers:

vaieri [72.5K]3 years ago

8 0

Answer:

12 should be 13 26 O on right

Explanation:

nikdorinn [45]3 years ago

6 0

Answer:

12 should be 13 26 O on right

Explanation:

only has 24 on left

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What is the balanced equation for the single replacement reaction between potassium and copper (2) chloride

frutty [35]

Answer: K + CuCl2 -> KCl2 + Cu

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4 years ago

Mario uses a hot plate to heat a beaker of 50mL of water. He used a thermometer to measure the temperature of the water. The wat

Alex777 [14]

Answer:

The boiling point decreases as the volume decreases.

Explanation:

The Temperature - Volume law otherwise called as Charles law is applied, which says that the volume of the given gas at constant pressure is directly proportional to the temperature measured in Kelvin. As the volume increases, the temperature also increases, if the volume decreases, then the temperature also decreases.

As per the Charles law, here the volume is decreased from 50 ml to 25 ml so the boiling point also decreases.

8 0

3 years ago

Read 2 more answers

If a sample contains 21.2 g N, how many moles of N does it contain?

cupoosta [38]

Hello!

If a sample contains 21.2 g N, how many moles of N does it contain?

0.66 mol

1.51 mol

14.01 mol

297.01 mol

We have the following data:

m (mass) = 21.2 g

MM (Molar mass of Nitrogen) = 14 g / mol

n (number of moles) = ?

Formula:

$n = \dfrac{m}{MM}$

Solving:

$n = \dfrac{m}{MM}$

$n = \dfrac{21.2\:\diagup\!\!\!\!\!g}{14\:\diagup\!\!\!\!\!g/mol}$

$\boxed{\boxed{n\approx 1.51\:mol}}\end{array}}\qquad\checkmark$

Answer:

1.51 mol

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

5 0

4 years ago

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What volume of 0.4567 m h2so4 is required to neutralize 30.00 ml of 0.3210 m naoh?

sertanlavr [38]

Answer is: volume of H₂SO₄ is 42.1 mL.
Chemical reaction: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
c(H₂SO₄) = 0,4567 M = 0,4567 mol/L.
V(NaOH) = 30 mL ÷ 1000 mL/L = 0,03 L.
c(NaOH) = 0,321 M = 0,321 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,321 mol/L · 0,030 L.
n(NaOH) = 0,00963 mol.
From chemical reaction: n(H₂SO₄) : n(NaOH) = 1 : 2.
n(H₂SO₄) = 0,01926 mol.
V(H₂SO₄) = n(H₂SO₄) ÷ c(H₂SO₄).
V(H₂SO₄) = 0,01926 mol ÷ 0,4567 mol/L.
V(H₂SO₄) = 0,0421 L = 42,1 mL.

3 0

3 years ago

What is the pH of a KOH solution that has [H ] = 1. 87 × 10–13 M? What is the pOH of a KOH solution that has [OH− ] = 5. 81 × 10

vlada-n [284]

pH is the hydrogen ion concentration and pOH is the hydroxide ion concentration in the solution. pH KOH is 12.73, pOH KOH is 2.24 and pH NaCl is 7.

pH is the negative log of the hydrogen ion concentration and pOH is the negative log of the hydroxide ion concentration.

The relation between the pH and pOH can be given as, $\rm pOH = 14 - pH$

The pH of KOH can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the first case, the concentration of the KOH is $1. 87 \times 10^{-13}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 87 \times 10^{-13}\;\rm M ]\\\\&= 12.73\end{aligned}$

Hence, the pH of KOH is 12.73.

pOH of KOH can be calculated by the formula,

$\rm pOH = \rm -log [OH^{-}]$

The hydroxide concentration of the KOH solution is $5. 81 \times 10^{-3}\;\rm M$

Substituting value in the equation:

$\begin{aligned} \rm pOH &= \rm -log [OH^{-}]\\\\&= \rm -log [5. 81 \times 10^{-3}\;\rm M ]\\\\&= 2.24 \end{aligned}$

Hence, the pOH of KOH is 2.24

The pH of NaCl can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the third case, the concentration of the NaCl is $1. 00\times 10^{-7}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 00 \times 10^{-7}\;\rm M ]\\\\&= 7 \end{aligned}$

Hence, the pH of KOH is 7.0.

Therefore, KOH is basic and NaCl is approximately neutral.

Learn more about pH and pOH here:

brainly.com/question/13885794

3 0

3 years ago