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Mariana [72]
3 years ago
15

What is the correct Lewis dot structure of NH2

Chemistry
1 answer:
Mrac [35]3 years ago
4 0
B I believe. Hope this helps you!


-Belle
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Calculate the volume (liters) of solution that can be prepared from 180. grams of K2Cr2O7 for a 0.200M solution.
Sladkaya [172]

Answer:

The volume of the solution is 0.305 liters.

Explanation:

Molar mass is the amount of mass that a substance contains in one mole. The molar mass of K₂Cr₂O₇ is 294 g / mole. Then you can apply the following rule of three: if by definition of molar mass 294 grams of the compound are contained in 1 mole, 180 grams are contained in how many moles?

moles=\frac{180 grams*1mole}{294 grams}

moles= 0.61

Molarity is a measure of the concentration of a substance that is defined as the number of moles contained in a certain volume. So, the molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

molarity=\frac{number of moles of solute}{volume}

Molarity is expressed in units \frac{moles}{liter}.

In this case:

  • molarity= 2 M
  • number of moles of solute= 0.61 moles
  • volume= ?

Replacing in the definition of molarity:

2 M=\frac{0.61 moles}{volume}

Solving:

volume=\frac{0.61 moles}{2 M}

volume= 0.305 liters

<u><em>The volume of the solution is 0.305 liters.</em></u>

8 0
2 years ago
What is the mass of silver chlorate (191.32 g/mol) that decomposes to release 0.466L of oxygen gas at STP? AgC1036) _AgCl). _026
vesna_86 [32]

Answer : The mass of silver chlorate will be 2.654 grams.

Explanation :

The balanced chemical reaction is,

2AgClO_3\rightarrow 2AgCl+3O_2

First we have to calculate the moles of oxygen gas at STP.

As, 22.4 L volume of oxygen gas present in 1 mole of oxygen gas

So, 0.466 L volume of oxygen gas present in \frac{0.466}{22.4}=0.0208 mole of oxygen gas

Now we have to calculate the moles of silver chlorate.

From the balanced chemical reaction, we conclude that

As, 3 moles of oxygen produced from 2 moles of silver chlorate

So, 0.0208 moles of oxygen produced from \frac{2}{3}\times 0.0208=0.01387 moles of silver chlorate

Now we have to calculate the mass of silver chlorate.

\text{Mass of }AgClO_3=\text{Moles of }AgClO_3\times \text{Molar mass of }AgClO_3

Molar mass of silver chlorate = 191.32 g/mole

\text{Mass of }AgClO_3=0.01387mole\times 191.32g/mole=2.654g

Therefore, the mass of silver chlorate will be 2.654 grams.

3 0
3 years ago
Why might an idea or hypothesis be discarded? A. If another scientist doesn't like it. B. If evidence also supports another hypo
vladimir1956 [14]

Answer:

C. If it is tested and the evidence does not support it.

Explanation:

A hypothesis is more less a scientific guess. Before such a guess or prediction is made, empirical observations and deductions are first made. It is from the result of the observations that a hypothesis statement is made.

For a hypothesis to become widely adopted and accepted, it must be testable within the limits of the experiment as described by the proposer. When subjected to test and it agrees, the status of a hypothesis can be upgraded.

If the hypothesis is tested and evidence contrasts the result being sort for, a hypothesis will be discarded.

7 0
3 years ago
O Previous
Ede4ka [16]

Answer:

what is scientific model

Explanation:

4 0
3 years ago
An unknown gaseous substance has a density of 1.06 g/L at 31 °C and 371 torr. If the substance has the following percent composi
Anarel [89]

Answer:

C) C4H6 - Right answer

Explanation:

Let's combine the Ideal Gases Law with density to get the molecular formula for the unknown gas.

Density = mass / volume

1.06 g /L means that 1.06 grams of compound occupy 1 liter of volume.

P . V = n . R . T

Pressure in Torr must be converted to atm

760 Torr are 1 atm

371 Torr  are __ (371 .1)/760 = 0.488 atm

0.488 atm . 1L = 1.06g/MM . 0.082 . 304K

(0.488 atm . 1L) / 0.082 . 304K = 1.06g/ MM

Mass / Molar mass = Moles → That's why the 1.06 g / MM

0.0195 mol = 1.06g / MM

1.06g/0.0195 mol = MM →  54.3 g/m

Now, let's use the composition

100 g of compound have 88.8 g of C

54.3 g of compound have ___ (54.3  . 88.8) /100 = 48 g of C

100 g of compound have 11.2 g of H

54.3 g of compound have __ (54.3  .  11.2)/100 = 6 g of H

48 g of C are included un 4 atoms

6 g of H are included in 6 atoms

4 0
3 years ago
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