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ivolga24 [154]
3 years ago
10

Determine whether the bonds in CCl4 and N2 are polar or non-polar. If the bond is polar, assign the partial positive and negativ

e charges. Electronegativity values: C = 2.5, Cl= 3.0, N= 3.0
Chemistry
1 answer:
stira [4]3 years ago
7 0

Refer to the attachment

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An object has a volume of 22 liters and a mass of 35 kilograms what the objects density? Will it float or sink in the water?
Kobotan [32]

Answer:

density = 1.59

it will sink in water

Explanation:

density = m/v

x = 35/22

x= 1.59

4 0
4 years ago
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Brrunno [24]

Answer:

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2. the other bulb would not light

3. the other bulb lit normally

4. the series curcuit with 1 bulb was brighter than the series circuit with 2 bulbs

5. both bulbs were illuminated equally

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7 0
3 years ago
Assuming all protons can be seen in this atom what is the charge in this ion
bija089 [108]

Answer:

A proton has positive charge of 1, that is, equal but opposite to the charge of an electron. A neutron, like the name implies, is neutral with no net charge. The charge is believed to be from the charge of the quarks that make up the nucleons (protons and neutrons). need a pictur of the question

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6 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
At a given temperature, the equilibrium constant for the formation of HI from H2 and I2 was found to be 29.9. Calculate the equi
evablogger [386]

Answer:

The correct answer is: K'= 0.033.

Explanation:

The formation of HI from H₂ and I₂ is given by:

H₂ + I₂ → 2 HI      K= 29.9

The decomposition of HI is the reverse reaction of the formation of HI:

2 HI → H₂ + I₂     K'

Thus, K' is the equilibrium constant for the reverse reaction of formation of HI. It is calculated as the reciprocal of the equilibrium constant of the forward reaction (K):

K' = 1/K = 1/(29.9)= 0.033

Therefore, the equilibrium constant for the decomposition of HI is K'= 0.033

5 0
4 years ago
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