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frozen [14]
2 years ago
9

Which type of reaction is NaOH + KNO3 NaNO3 + KOH -->>

Chemistry
1 answer:
Artyom0805 [142]2 years ago
4 0

Answer:

Explanation:

NaOH + KNO3 --> NaNO3 + KOH. combustion. CH4 + 2 O2 --> CO2 + 2 H2O. single displacement. 2 Fe + 6 NaBr --> 2 FeBr3 + 6 Na. double displacement.

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What type of compound is calcium oxide
DIA [1.3K]
A carbonate because the oxygen neutralizes the reactive coding of the calcium in its original form.
5 0
3 years ago
Which species is the conjugate acid of the hemiacetal formed by reaction of benzaldehyde with methanol containing a trace of aci
fiasKO [112]

Answer:

Sulphuric acid is the on

Explanation:

think so

3 0
3 years ago
A sample containing 16.93 metal pellets is poured into a graduated cylinder containing 11.2 of water, causing the water level in
mixer [17]

Answer:

<h3>The answer is 1.99 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 16.93 g

volume = final volume of water - initial volume of water

volume = 19.7 - 11.2 = 8.5 mL

We have

density =  \frac{16.93}{8.5}  \\  = 1.9917647058...

We have the final answer as

<h3>1.99 g/mL</h3>

Hope this helps you

7 0
3 years ago
Calcium dihydrogen phosphate, Ca(H₂PO₄)₂, and sodium hydrogen carbonate, NaHCO₃, are ingredients of baking powder that react to
NikAS [45]

0.012 mol of CO₂ can be produced from 3.50 g of baking powder.

<h3>What is baking powder?</h3>
  • Baking powder is a dry chemical leavener composed of carbonate or bicarbonate and a weak acid.
  • The addition of a buffer, such as cornstarch, prevents the base and acid from reacting prematurely.
  • Baking powder is used in baked goods to increase volume and lighten the texture.

To find how many moles of CO₂ are produced from 1.00 g of baking powder:

The balanced equation is:

  • Ca(H₂PO₄)₂(s) + 2NaHCO₃(s) → 2CO₂(g) + 2H₂O(g) + CaHPO₄(s) + Na₂HPO₄(s)

On 3.50 g of baking power:

  • mCa(H₂PO₄)₂ = 0.35 × 3.50 = 1.225 g
  • mNaHCO₃ = 0.31 × 3.50 = 1.085 g

The molar masses are: Ca = 40 g/mol; H = 1 g/mol; P = 31 g/mol; O = 16 g/mol; Na = 23 g/mol; C = 12 g/mol.

So,

  • Ca(H₂PO₄)₂: 40 + 4 × 1 + 31 + 8 × 16 = 203 g/mol
  • NaHCO₃: 23 + 1 + 12 + 3 × 16 = 84 g/mol

The number of moles is the mass divided by molar mass, so:

  • nCa(H₂PO₄)₂ = 1.225/203 = 0.006 mol
  • nNaHCO₃ = 1.085/84 = 0.0129 mol

First, let's find which reactant is limiting.

Testing for Ca(H₂PO₄)₂, the stoichiometry is:

  • 1 mol of Ca(H₂PO₄)₂ ---------- 2 mol of NaHCO₃
  • 0.006 of Ca(H₂PO₄)₂ -------- x

By a simple direct three rule:

  • x = 0.012 mol

So, NaHCO₃ is in excess.

The stoichiometry calculus must be done with the limiting reactant, then:

  • 1 mol of Ca(H₂PO₄)₂ ------------- 2 mol of CO₂
  • 0.006 of Ca(H₂PO₄)₂ -------- x

By a simple direct three rule:

  • x = 0.012 mol of CO₂

Therefore, 0.012 mol of CO₂ can be produced from 3.50 g of baking powder.

Know more about baking powder here:

brainly.com/question/20628766

#SPJ4

The correct question is given below:

Calcium dihydrogen phosphate, Ca(H2PO4)2, and sodium hydrogen carbonate, NaHCO3, are ingredients of baking powder that react with each other to produce CO2, which causes dough or batter to rise: Ca(H2PO4)2(s) + NaHCO3(s) → CO2(g) + H2O(g) + CaHPO4(s) + Na2HPO4(s)[unbalanced] If the baking powder contains 31.0% NaHCO3 and 35.0% Ca(H2PO4)2 by mass: (a) How many moles of CO2 are produced from 3.50 g of baking powder?

3 0
2 years ago
For the reaction of hydrogen with iodine
fenix001 [56]

Answer:

r_{H_2} = \frac{-1}{2} r_{HI}

Explanation:

Hello!

In this case, considering the given chemical reaction:

H_2(g) + I_2(g) \rightarrow 2HI(g)

Thus, by applying the law of rate proportions, we can write:

\frac{1}{-1} r_{H_2} = \frac{1}{-1}r_{i_2} = \frac{1}{2} r_{HI}

Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:

r_{H_2} = \frac{-1}{2} r_{HI}

Best regards!

8 0
2 years ago
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