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Ede4ka [16]
3 years ago
13

Which of the following would be produced in a combustion reaction?

Chemistry
2 answers:
Vaselesa [24]3 years ago
7 0
Carbon dioxide I believe I’m sorry if u do get it wrong
Free_Kalibri [48]3 years ago
4 0

Answer:

C. Carbon Dioxide because combustion means to blow up, therefor starting a fire, and fire produces Carbon Dioxide.

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Which of the following contains the greatest mass of oxygen?
Art [367]

Answer:

letter d

Explanation:

hope it's help you

5 0
3 years ago
When a solid is heated up what occurs? A. The molecules move faster and become more dense B. The molecules move faster and becom
Xelga [282]

Answer: B. The molecules move faster and become less dense

Explanation:

Liquid is the state of matter in which particles are less tightly bound as compared to solids as they have weak inter molecular forces as compared to solids.

Thus when a solid is heated, the molecules start moving in random motion due to an increase in the kinetic energy. The molecules move apart as the forces will be weak and thus less molecules will be present per unit volume. Thus it will become less dense.

5 0
3 years ago
100 points! Brainliest if correct!!
Harman [31]
<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ 4 x 2 = 8

4 x 1 = 4

4 x 4 = 16

16 + 8 + 4 = 28

The answer is option D. 28

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

4 0
3 years ago
Read 2 more answers
Write the equilibrium constant expression for this reaction: 2H+(aq)+CO−23(aq) → H2CO3(aq)
MrRissso [65]

Answer:

Equilibrium constant expression for \rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq) \rightleftharpoons H_2CO_3\, (aq):

\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{[\mathrm{H_2CO_3}]}{\left[\mathrm{H^{+}\, (aq)}\right]^{2} \, \left[\mathrm{CO_3}^{2-}\right]}.

Where

  • a_{\mathrm{H_2CO_3}}, a_{\mathrm{H^{+}}}, and a_{\mathrm{CO_3}^{2-}} denote the activities of the three species, and
  • [\mathrm{H_2CO_3}], \left[\mathrm{H^{+}}\right], and \left[\mathrm{CO_3}^{2-}\right] denote the concentrations of the three species.

Explanation:

<h3>Equilibrium Constant Expression</h3>

The equilibrium constant expression of a (reversible) reaction takes the form a fraction.

Multiply the activity of each product of this reaction to get the numerator.\rm H_2CO_3\; (aq) is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be \left(a_{\mathrm{H_2CO_3\, (aq)}}\right).

Similarly, multiply the activity of each reactant of this reaction to obtain the denominator. Note the coefficient "2" on the product side of this reaction. \rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq) is equivalent to \rm H^{+}\, (aq) + H^{+}\, (aq) + {CO_3}^{2-}\, (aq). The species \rm H^{+}\, (aq) appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:

\left(a_{\mathrm{H^{+}}}\right)\cdot \left(a_{\mathrm{H^{+}}}\right)\cdot \, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right = \left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right.

That's where the exponent "2" in this equilibrium constant expression came from.

Combine these two parts to obtain the equilibrium constant expression:

\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \quad\begin{matrix}\leftarrow \text{from products} \\[0.5em] \leftarrow \text{from reactants}\end{matrix}.

<h3 /><h3>Equilibrium Constant of Concentration</h3>

In dilute solutions, the equilibrium constant expression can be approximated with the concentrations of the aqueous "(\rm aq)" species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:

\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{\left[\mathrm{H_2CO_3\, (aq)}\right]}{\left[\mathrm{H^{+}\, (aq)}\right]^2\cdot \left[\mathrm{{CO_3}^{2-}\, (aq)}\right]}.

8 0
3 years ago
How many grams of NaOH is needed to neutralize 90 mL of 1.5 N HCl?
AnnZ [28]

mol = conc × v

= 1.5 × 0.09

= 0.135 moles of HCl

HCl + NaOH > NaCl + H2O

1 mole HCl = 1 mole NaOH

0.135 mol HCl = x

x = 0.135 mol NaOH

mass = mol × molar mass

= 0.135 × 40

= 5.4 g

NaOH = 23 + 16 + 1 = 40 g/mol

I'm not a 100% sure if it's correct

6 0
3 years ago
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