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3 years ago

Which of the following statements describes the correct method of preparation of 1.00 L of a 2.0 M urea solution?

Chemistry

1 answer:

Zolol [24]3 years ago

8 0

Answer:

To prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution

Explanation:

Molarity of a solute in a solution denotes number of moles of solute dissolved in 1 L of solution.

So, moles of urea in 1.00 L of a 2.0 M urea solution = 2 moles

We know, number of moles of a compound is the ratio of mass to molar mass of that compound.

So, mass of 2 moles of urea = $(2\times 60.06)g=120 g$

Therefore to prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution

So, option (C) is correct.

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Which ecosystem is the least stable?

Softa [21]

Answer:

tundra bc it has barley any food source and has vary little inhabitants

Explanation:

6 0

3 years ago

Un átomo X posee 29 protones y de carga +2 ¿Cuántos electrones tiene?

lesya [120]

Responder:

Explicación:

Dado que:

Número de protones en el átomo X = 29

Carga en el átomo X = +2

Si no hay cargo neto;

número de protones = número de electrones

Sin embargo, dado que el átomo X tiene una carga de +2 (dando 2 electrones).

Por lo tanto,

Número de electrones = número de protones - número de carga en el átomo)

Número de electrones = (29 - 2) = 27

4 0

3 years ago

Dustin is mixing concrete. A formula for concrete calls for 3/8 gal of water. Dustin wants to make 7/9 more concrete than the fo

slava [35]

We can use a ratio to solve this question. Lets refer to the amount formed by the formula as a serving
1 serving needs 3/8 gal
We want to create 7/9 servings extra, so
1 + 7/9 = 16/7
1 : 3/8
16/7 : x
1/(3/8) = (16/7) / x
8 / 3 = 16x / 7
x = 7/6
He needs to use 7/6 gallons of water.

5 0

3 years ago

2074 Set B Q.No. 1 What mass of nitrogen will be requires

timurjin [86]

140 g of nitrogen (N₂)

Explanation:

We have the following chemical equation:

N₂ + 3 H₂ -- > 2 NH₃

Now, to find the number of moles of ammonia we use the Avogadro's number:

if 1 mole of ammonia contains 6.022 × 10²³ molecules

then X moles of ammonia contains 6.022 × 10²⁴ molecules

X = (1 × 6.022 × 10²⁴) / 6.022 × 10²³

X = 10 moles of ammonia

Taking in account the chemical reaction we devise the following reasoning:

If 1 mole of nitrogen produces 2 moles of ammonia

then Y moles of nitrogen produces 10 moles of ammonia

Y = (1 × 10) / 2

Y = 5 moles of nitrogen

number of moles = mass / molecular weight

mass = number of moles × molecular weight

mass of nitrogen (N₂) = 5 × 28 = 140 g

Learn more about:

Avogadro's number

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#learnwithBrainly

7 0

3 years ago

A sample of gas has a density of 0.53 g/L at 225 K and under a pressure of 108.8 kPa. Find the density of the gas at 345 K under

sukhopar [10]

Answer:

$\rho _2=0.22g/L$

Explanation:

Hello!

In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

$PV=nRT\\\\PV=\frac{m}{MM} RT\\\\P*MM=\frac{m}{V} RT\\\\P*MM=\rho RT$

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

$\frac{P_1*MM}{P_2*MM} =\frac{\rho _1RT_1}{\rho _2RT_2} \\\\\frac{P_1}{P_2} =\frac{\rho _1T_1}{\rho _2T_2}$

It means we can compute the final density as shown below:

$\rho _2=\frac{\rho _1T_1P_2}{P_1T_2}$

Now, we plug in to obtain:

$\rho _2=\frac{0.53g/L*225K*68.3kPa}{345K*108.8kPa}\\\\\rho _2=0.22g/L$

Regards!

8 0

2 years ago