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3 years ago

What is the total number of moles of solute contained in 0.5 liter of 3.0 m

Chemistry

1 answer:

myrzilka [38]3 years ago

4 0

Based on science 4) 1) 0.25M

2) 0.66M

3) 1.5M

4) 4.0M ––> Therefore, this is the most concentrated

5) 30g NaOH(1 mol NaOH / 40gNaOH) = 0.75 mol NaOH / 0.500L = 1.5M NaOH (3)

6) 1) 1.0 mol / L(1 L) = 1 mol H2SO4

2) 1 mol / L (2 L) = 2 mol H2SO4

3) 0.50 mol / L ( 1.0L) = 0.50 mol H2SO4 ––>

7) 0.200 mol / L (1 L)(74.6g / mol) = 14.92g (2)

8) 0.25 mol / 0.250 L = 1 mol / L = 1M (1)

This is the answer : 1m

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Jamal is working with three ionic compounds: sodium chloride, calcium sulfide, and barium oxide. His teacher asks him which are

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Answer:

option A is correct ( sodium, calcium and barium)

Explanation:

Given compounds:

Sodium chloride , calcium sulfide, barium oxide

We know that metals form positive ions. In order to solve the problem we must identify the metals from given compounds.

Na⁺Cl⁻

Ca²⁺S²⁻

Ba²⁺O²⁻

We can see that sodium, calcium and barium contain positive charges.

Thus option A is correct.

Because sodium have one valance electron. When it combine with chlorine sodium lose its one electron to complete the octet and chlorine accept it to complete its octet. Thus sodium form positive ion and chlorine form negative ion.

Similarly barium and calcium are present in group 2. Both have two valance electron. When they lose them cation are formed.

Other option are incorrect because,

Option B have sulfur and oxygen which are anion.

Option C have chlorine which is also anion

Option D have chlorine, sulfur and oxygen that are anions.

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3 years ago

Penicillinase, also known as β‑lactamase, is a bacterial enzyme that hydrolyzes and inactivates the antibiotic penicillin. Penic

sergeinik [125]

Explanation:

The given data is as follows.

$V_{max} = 6.8 \times 10^{-10} \mu mol/min$

$K_{m} = 5.2 \times 10^{-6} M$

Now, according to Michaelis-Menten kinetics,

$V_{o} = V_{max} \times [\frac{S}{(S + Km)}]$

where, S = substrate concentration = $10.4 \times 10^{-6}$ M

Now, putting the given values into the above formula as follows.

$V_{o} = V_{max} \times [\frac{S}{(S + Km)}]$

$V_{o} = 6.8 \times 10^{-10} \mu mol/min \times [\frac{10.4 \times 10^{-6} M}{(10.4 \times 10^{-6}M + 5.2 \times 10^{-6} M)}]$

$V_{o} = 6.8 \times 10^{-10} \mu mol/min \times 0.667$

= $4.5 \times 10^{-10} \mu mol/min$

This means that $V_{o}$ would approache $V_{max}$ .

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3 years ago

Which compound contains both ionic and covalent

musickatia [10]

Answer:

(3) NaNO₃

Step-by-step explanation:

Sodium nitrate has ionic bonds, because it consists of Na⁺ and NO₃⁻ ions.

However, the nitrate ions have covalent bonds between the O atoms and the central N atoms.

(1) and (2) are wrong. Both N₂O₅ and HCl consist of nonmetals, so they are covalent compounds.

(4) is wrong. NaCl has only ionic bonds between the Na⁺ and Cl⁻ ions

7 0

3 years ago

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A 720. cm^3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.966 g at 25.0°C and the pressure is 760.

sleet_krkn [62]

Explanation:

The given data is as follows.

Pressure (P) = 760 torr = 1 atm

Volume (V) = $720 cm^{3}$ = 0.720 L

Temperature (T) = $25^{o}C$ = (25 + 273) K = 298 K

Using ideal gas equation, we will calculate the number of moles as follows.

PV = nRT

Total atoms present (n) = $\frac{PV}{RT}$

= $1 \times \frac{0.720 L}{0.0821 \times 298}$

= 0.0294 mol

Let us assume that there are x mol of Ar and y mol of Xe.

Hence, total number of moles will be as follows.

x + y = 0.0294

Also, 40x + 131y = 2.966

x = 0.0097 mol

y = (0.0294 - 0.0097)

= 0.0197 mol

Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = $\frac{y}{(x+y)}$

= $\frac{0.0197}{0.0294}$

= 0.67

Therefore, the mole fraction of Xe is 0.67.

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3 years ago

Iron and oxygen form rust. How many moles of rust should be produced if 1

kakasveta [241]

Answer:

1.387 moles

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

4Fe + 3O2 —> 2Fe2O3

Step 2:

Determination of the number of mole of Fe in 155.321g of Fe. This can be achieved by doing the following:

Mass of Fe = 155.321g

Molar Mass of Fe = 56g/mol

Number of mole of Fe =?

Number of mole = Mass/Molar Mass

Number of mole of Fe = 155.321/56

Number of mole of Fe = 2.774 mol

Step 3:

Determination of the number of mole of rust (Fe2O3) produced. This is illustrated below:

From the balanced equation above,

4 moles of Fe produced 2 moles of Fe2O3.

Therefore, 2.774 moles of Fe will produce = (2.774 x 2)/4 = 1.387 moles of Fe2O3.

Therefore, 1.387 moles of rust (Fe2O3) is produced from the reaction

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3 years ago

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