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5 days ago

What is the stoichiometric ratio between BaCl2 and NaCl

1 answer:

7 0

<span>BaCl2 + Na2SO4 --> BaSO4 + 2NaCl In this reaction, 1.0 g of BaCl2 and 1.0 g of Na2SO4 are present. We need to identify the limiting reactant. "First, convert grams to moles" 1.0 g BaCl2 * (1 mol BaCl2 / 208.2 g BaCl2) = 4.8 x 10^-3 mol BaCl2 1.0 g Na2SO4 * (1 mol Na2SO4 / 142.04 g Na2SO4) = 7.0 x 10^-3 mol Na2SO4 (7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2) = 1.5 mol Na2SO4 per mol BaCl2 "Using this ratio to compare with the balanced equation, BaCl2 + Na2SO4 --> BaSO4 + 2NaCl" The balanced equation indicates that 1 mol of BaCl2 reacts with 1 mol of Na2SO4. However, we found that 1.5 mol of Na2SO4 is available for each mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>

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A tritium nucleus is formed by combining two neutrons and a proton. the mass of this nucleus is 9.106 × 10–3 universal mass unit

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E = mc²

where E = energy produced

m = mass of the nucleus

C = speed of light

m = 9.106 x 10⁻³ x 1.67 x 10⁻²⁷ kg

C = 3 x 10⁸ m/s, thus C² = 9 x 10¹⁶

E = 1.37 x 10⁻¹² J

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5 days ago

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The next 3 questions will walk you through using the Henderson-Hasselbalch equation for the following question. For each step pr

lions [1003]

Response:

The pKa value is 13.0.

Clarification:

pKa + pKb = 14

For trimethylamine, Kb = 6.3 × $10^{-5}$

Calculating pKb: pKb = - log (6.3 × $10^{-5}$ )

= 1.0

Thus, pKa = 14 - pKb = 14 - 1.0

pKa = 13.0

Verification: The typical range for pKa in weak acids is from 2 to 13.

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5 days ago

The concentration of Si in an Fe-Si alloy is 0.25 wt%. What is the concentration in kilograms of Si per cubic meter of alloy?

KiRa [976]

Answer: The mass of Si in kilograms is, $19.55kg/m^3$

Explanation:

Given that the Si concentration in an Fe-Si alloy is 0.25 weight percent, this translates to:

Mass of Si = 0.25 g = 0.00025 kg

Mass of Fe = 100 - 0.25 = 99.75 g = 0.09975 kg

Density of Si = $2.32g/cm^3=2.32\times 10^6g/m^3$

Density of Fe = $7.87g/cm^3=7.87\times 10^6g/m^3$

Next, we need to find the quantity of Si in kilograms per cubic meter of alloy.

Si concentration in kilograms = $\frac{\text{Weight of Si in 100 g of alloy}}{\text{Volume of 100 g of alloy}}$

Si concentration in kilograms = $\frac{\text{Weight of Si in 100 g of alloy}}{\frac{\text{Wight of Fe}}{\text{Density of Fe}}+\frac{\text{Wight of Si}}{\text{Density of Si}}}$

By substituting all the provided values into this formula, we arrive at:

Si concentration in kilograms = $\frac{0.00025kg}{\frac{99.75g}{7.87\times 10^6g/m^3}+\frac{0.25g}{2.23\times 10^6g/m^3}}$

Si concentration in kilograms = $19.55kg/m^3$

Hence, the mass of Si in kilograms is, $19.55kg/m^3$

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8 days ago

One tank of gold fish is fed the normal amount of food once a day. A second tank is fed twice a day. A third tank is fed four ti

Tems11 [854]

The inquiry is incomplete; here is the full question:

One tank of goldfish receives the standard amount of feeding once daily, a second tank is given two feedings a day, and a third tank is fed four times daily throughout a six-week experiment. The body fat of the fish is recorded every day.

Independent Variable-

Dependent Variable-

Constants

Control Group-

Answer:

A) The quantity of food given to the goldfish

B) The body fat of the goldfish

C) -Type of fish in the experiment (goldfish)

Time period for feeding the fish (six weeks)

Shape and size of the tanks

D) group of goldfish receiving the standard feeding amount

Explanation:

The objective of the experiment is to assess how the quantity of food affects the body fat of goldfish. Consequently, the amount of food serves as the independent variable while the body fat acts as the dependent variable.

The control group is the one given the standard feeding amount (once daily). All subjects are goldfish, fed over a six-week duration, with all tanks being the same shape and size, establishing the constants in the research.

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15 days ago

The following five beakers, each containing a solution of sodium chloride (NaCl, also known as table salt), were found on a lab

Anarel [852]

Answer:

Please review the following responses

Explanation:

1) A solution of 100. mL contains 19.5 g of NaCl (3.3M)

2) 100. mL of NaCl solution at 3.00 M (3 M)

3) A solution of 150. mL holds 19.5 g of NaCl (2.2 M)

4) The concentrations of beakers 1 and 5 are identical (1.5M)

Molar mass of NaCl = 23 + 36 = 59 g

For beaker number 3:

59 g -------------- 1 mol

19.5 g ------------- x

x = 19.5 x 1/59 = 0.33 mol

Molarity (M) = 0.33 mol/0.150 l = 2.2 M

For beaker number 4:

Molarity (M) = 0.33mol/0.10 l = 3.3 M

For beaker number 5:

Molarity (M) = 0.450/0.3 = 1.5 M

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7 days ago

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