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3 years ago

8

Cho hai điện tích q1 = 2.10-6 C, q2 = 4. 10-6C, đặt tại A và B trong không khí biết AB = 2 cm. Xác định vectơ cường độ điện trườ

ng tại:
a) H, là trung điểm của AB.
b) . M, MA = 1 cm, MB = 3 cm.
c) N, biết rằng NAB là một tam giác đều.

1 answer:

Nitella [24]3 years ago

6 0

Answer:

<h3>A</h3>

Explanation:

<h2>sorry try lang baka mali</h2><h2 />

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STALIN [3.7K]

What should the pharmacy technician include on the label if the mix is a multidose container and not all of the medication was used? The patient's name, in case he or she returns for another dose at a later time b) The technician's initials, date and time of preparation, expiration date of the medication, and its final strength c) Other afflictions the medication may assist with for future use d) Instructions on how and where to store the medication

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3 years ago

Calculate the mass of 6.9 moles of nitrous acid (HNO2). Explain the process or show your work by including all values used to de

AlexFokin [52]

Answer:- 324.3 grams.

Solution:- We have been given with 6.0 moles of nitrous acid and asked to calculate it's grams. Moles to grams is a unit conversion and for doing this conversion we multiply the given moles by the molar mass of the compound.

Molar mass is the formula mass and to calculate this the atomic masses of each atom are multiplied by their respective subscripts that is the number of the atom in the compound .

For example, $NO_2$ has one nitrogen atom and one oxygen atom. So, the molar mass of this is = atomic mass of N + 2(atomic mass of O)

= 14 + 2(16)

= 14 + 32

= 46 gram per mol

gram per mol is the unit of molar mass. So, the molar mass of $NO_2$ is 46 grams per mol.

Let's calculate the molar mass of nitrous acid using the same concept.

molar mass of [ $HNO_2$ = 1 + 14 + 2(16)

= 1 + 14 + 32

= 47 grams per mol

Now, 6.9 moles of nitrous acid could easily be converted to grams as:

$6.9molHNO_2(\frac{47g}{1mol})$

= 324.3 g

Hence, the mass of 6.9 moles of nitrous acid is 324.3 grams.

6 0

3 years ago

1) A 10g sample of H2(g) reacts with a 22g sample of O2(g) according to

solmaris [256]

Answer:

H₂ is excess reactant and O₂ the limiting reactant

Explanation:

Based on the chemical reaction:

2H₂(g) + O₂(g) → 2H₂O

<em>2 moles of H₂ react per mole of O₂</em>

<em />

To find limiting reactant we need to convert the mass of each reactant to moles:

<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>

10g H₂ * (1mol / 2.016g) = 4.96 moles

<em>Moles O₂ -Molar mass: 32g/mol-:</em>

22g O₂ * (1mol / 32g) = 0.69 moles

For a complete reaction of 0.69 moles of O₂ are needed:

0.69mol O₂ * (2mol H₂ / 1mol O₂) = 1.38 moles of H₂

As there are 4.96 moles,

<h3>H₂ is excess reactant and O₂ the limiting reactant</h3>

7 0

4 years ago

What happens to a gas as it is cooled below 0*C

Damm [24]

Sorry but it depends on the gas in question and it's temperature of condensing /boiling point if this it 0° it will turn liquid

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4 years ago

Read 2 more answers

odium carbonate (Na2CO3Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be add

Arte-miy333 [17]

Answer : The mass of sodium carbonate added to neutralize must be, $6.54\times 10^3kg$

Explanation :

First we have to calculate the moles of $H_2SO_4$ .

$\text{Moles of }H_2SO_4=\frac{\text{Mass of }H_2SO_4}{\text{Molar mass of }H_2SO_4}$

Given:

Molar mass of $H_2SO_4$ = 98 g/mole

Mass of $H_2SO_4$ = $6.05\times 10^3kg=6.05\times 10^6g$

Conversion used : (1 kg = 1000 g)

Now put all the given values in the above expression, we get:

$\text{Moles of }H_2SO_4=\frac{6.05\times 10^6g}{98g/mol}=6.17\times 10^4mol$

The moles of $H_2SO_4$ is, $6.17\times 10^4mol$

Now we have to calculate the moles of $Na_2CO_3$

The balanced neutralization reaction is:

$Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2CO_3$

From the balanced chemical reaction we conclude that,

As, 1 mole of $H_2SO_4$ neutralizes 1 mole of $Na_2CO_3$

So, $6.17\times 10^4mol$ of $H_2SO_4$ neutralizes

Now we have to calculate the mass of $Na_2CO_3$

$\text{ Mass of }Na_2CO_3=\text{ Moles of }Na_2CO_3\times \text{ Molar mass of }Na_2CO_3$

Molar mass of $Na_2CO_3$ = 106 g/mole

$\text{ Mass of }Na_2CO_3=(6.17\times 10^4mol)\times (106g/mole)=6.54\times 10^6g=6.54\times 10^3kg$

Thus, the mass of sodium carbonate added to neutralize must be, $6.54\times 10^3kg$

3 0

4 years ago