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3 years ago

What is the common difference between successive terms in the sequence? 9, 2.5,-4, -10.5,-17,

Chemistry

2 answers:

Vlad [161]3 years ago

8 0

Answer:

-6.5

Explanation:

Reil [10]3 years ago

7 0

6.5 is subtracted from each number.
9 - 6.5 = 2.5
2.5 - 6.5 = -4
-4 - 6.5 = -10.5
-10.5 - 6.5 = -17
or you could think of adding -6.5 to everything if that is easier for you to picture.

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Which chemical equation correctly represents the synthesis of calcium sulfite from calcium oxide and sulfur dioxide ?

Helen [10]

Answer:

haggis is the best place for a new dog and what's going to happen with the

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3 years ago

For 100.0 mL of a solution that is 0.040M CH3COOH and 0.010 M CH3COO, what would be the pH after adding 10.0 mL 50.0 mM HCl?

damaskus [11]

Answer:

The pH of the buffer is 3.90

Explanation:

The mixture of a weak acid CH3COOH and its conjugate base CH3COO produce a buffer that follows the equation:

pH = pKa + log [A-] / [HA]

Where pH is the pH of the buffer, pKa is the pKa of acetic acid (4.75), and [A-] could be taken as the moles of the conjugate base and [HA] the moles of thw weak acid.

To solve this question we need to find the moles of the CH3COOH and CH3COO- after the reaction with HCl:

CH3COO- + HCl → CH3COOH + Cl-

The moles of CH3COO- are its initial moles - the moles of HCl added

And moles of CH3COOH are its initial moles + moles HCl added

Moles CH3COO-:

Initial moles = 0.100L * (0.010mol / L) = 0.00100moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.000500 moles

Moles CH3COOH:

Initial moles = 0.100L * (0.040mol / L) = 0.00400moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.003500 moles

pH is:

pH = 4.75 + log [0.000500] / [0.00350]

pH = 3.90

<h3>The pH of the buffer is 3.90</h3>

3 0

3 years ago

Which part of metallic bonding theory helps the most to explain why metals are able to conduct electricity?

devlian [24]

Explanation:

The metallic bonding theory basically states that valence electrons are free, delocalized, mobile, and not associated with any particular atom. It is pften regarded as a sea of electrons surrounding te atoms.

Metals are able to conduct electricity because of the free movement of the electrons (they are not bounded to a particular atom). Hence, electricity can be defined as the movement of electrons.

3 0

3 years ago

how many moles of hydrogen are evolved when an excess of sodium metal is added to one mole of citric acid?

vova2212 [387]

The formula of citric acid is:

   CO2H
   |
HO - C - CH2CO2H
   |
C - CH2 CO2H

I bolded the 4 hydrogens that can be subsituted by sodium.

Three of those hydrogens belong to carboxyl group (- CO2H) and one to a hydroxyl group (- OH).

You need to take into account that 4 H will form 2 H2 molecules.

So, the answer is that 2 moles of molecules will be formed from 1 mole of citric acid molecules.

4 0

3 years ago

1.16g of methane burns completely with 4.16g of oxygen to form 3.52g of carbon dioxide and

Klio2033 [76]

Answer:

weight of H₂O formed = 2.3 grams H₂O (2 sig. figs.)

Explanation:

Rxn: CH₄(g) + 2O₂(g) => CO₂(g) + 2H₂O(g)

Given: 1.16g 4.16g 3.52g ? (g)

Moles: 1.16g/16g·mol⁻¹ 4.16g/32g·mol⁻¹ 3.52g/44g·mol⁻¹

= 0.0725mol. = 0.13mol. = 0.08mol. => ? (moles)

Limiting Reactant: Divide each mole value by related coefficient of balanced standard equation (that is, balanced with coefficients in lowest whole number ratios). The smaller value is the limiting reactant.

0.0725/1 0.13/2 0.080/1

= 0.0725 = 0.065 = 0.080

Limiting Reactant is O₂(g) => 0.065 is smaller value after dividing each mole value by related coefficient of balanced equation.

NOTE: When working problem, however, one must use the mole value calculated from given amount in grams. That is, in this case 0.13 mole O₂. The 'divide by related coefficient and check smaller value' is ONLY for identifying the limiting reactant. This trick works for ALL general chemistry problems.

Moles H₂O formed: Since the coefficient of the limiting reactant (O₂) equals the coefficient of water (H₂O), then the moles of water formed is 0.065 mole H₂O.

Weight (in grams) of H₂O formed:

Grams H₂O = moles H₂O x formula weight H₂O

= 0.13 mole H₂O x 18 g H₂O/mole H₂O

= 2.34 g H₂O (calculator answer)

= 2.3 g H₂O (final answer should be rounded to 2 sig. figs.) => form of final answer should be based on data in final computation having the least number of sig. figs.

Review: Sequence of calculations

Write and balance equation to smallest whole no. ratio of coefficients.
If not in moles, convert given 'measured' data to dimension of moles. => moles = mass (g)/formula wt(g·mol⁻¹) => moles = volume of gas in Liters/Std Molar Volume (= 22.4L·mole⁻¹ at STP) => moles = no. of particles / Avogadro's No. (= 6.02 x 10²³ part's/mole)
Determine Limiting Reactant => mole values of each compound given / related coefficient in standard equation => smallest value is L.R.
Determine moles of unknown needed/used/formed from limiting reactant in moles and coefficient of unknown compound in standard equation given data values.

=> moles of limiting reactant / coefficient of same cpd. in std. equation = unknown (X) / coefficient of same (unknown) cpd. in std. equation

=> cross multiply and solve for unknown (X)

=> L.R.(moles) / eqn. coef. of L.R. = X / eqn. coef. of X

=> (L.R.(calc'd moles)(eqn. coef. of X) = (X)(eqn. coef. of L.R.)

=> X (in moles) = (L.R.(calc'd. moles)(eqn. coef. of X) / (eqn. coef. of L.R.)

Convert X-answer in moles to desired dimension specified in problem.

=> grams = moles x formula wt.

=> volume (L) = moles x std. volume (= 22.4L/mole)

=> #particles = moles x Avogadro's Number (= 6.02 x 10²³ parts/mole)

8 0

3 years ago