Ask question

Ask question

All categories

3 years ago

8

How many mL of 0.40 M barium chloride are needed to react 175 mL of 1.50 M ammonium phosphate in the precipitation of barium pho

sphate?

1 answer:

timurjin [86]3 years ago

7 0

Answer:

656.25 mL

Explanation:

M1 = 0.40 M

M2 = 1.50 M

V2 = 175 mL

Plug those values into the equation:

M1V1 = M2V2

(0.40)V1 = (1.50)(175)

---> V1 = 656.25 mL

You might be interested in

The amount of gravitational force impacts the weight of objects.<br> true or false

ElenaW [278]

Answer:

True

Explanation:

It's how gravity works.

8 0

3 years ago

How many grams of fluorine must be reacted with excess lithium iodide to produce 10.0 grams of lithium fluoride?

rewona [7]

Answer:
7.32 g of F₂

Solution:
The equation is as follow,

2 LiI + F₂ → 2 LiF + I₂

According to equation,

51.88 g (2 mole) of LiF is produced from = 37.99 g (1 mole) F₂
So,
10 g of LiF will be produced by = X g of F₂

Solving for X,
X = (10 g × 37.99 g) ÷ 51.88 g

X = 7.32 g of F₂

8 0

4 years ago

Particles of matter how have both potential and kinetic energy is that true or false?

guapka [62]

False: No,any particles of matter do not have any potential or kinetic energy.

5 0

4 years ago

A synthesis reaction takes place when carbon monoxide (CO) and hydrogen gas (H2) react to form methanol (CH3OH). How many grams

jek_recluse [69]

The mass of methanol produced is 8.0 g.

We have the masses of two reactants, so this is a <em>limiting reactant</em> problem.

We know that we will need a <em>balanced equation</em> with masses, moles, and molar masses of the compounds involved.

<em>Step 1</em>. <em>Gather all the informatio</em>n in one place with molar masses above the formulas and everything else below them.

MM: ___28.01 2.016 ___32.04

_______CO + 2H_2 → CH_3OH

Mass/g: 7.0 __2.5

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>

Moles of CO = 7.0 g CO × (1 mol CO/28.01g CO) = 0.250 mol CO

Moles of H_2 =2.5 g H_2 × (1 mol H_2/2.016 g H_2) = 1.24 mol H_2

<em>Step 3. </em>Identify the<em> limiting reactan</em>t

Calculate the <em>moles of CH_3OH</em> we can obtain from each reactant.

<em>From CO</em>: Moles of CH_3OH = 0.250 mol CO × (1 mol CH_3OH /1 mol CO)

= 0.250 mol CH_3OH

<em>From H_2</em>: Moles of CH_3OH = 1.24 mol H_2 × (1 mol CH_3OH /2 mol H_2)

= 0.620 mol CH_3OH

<em>CO is the limiting reactant</em> because it gives the smaller amount of CH_3OH.

<em>Step 4</em>. Calculate the <em>mass of CH_3OH</em>

Mass of CH_3OH = 0.250 mol CH_3OH × (32.04 g CH_3OH /1 mol CH_3OH) = 8.0 g CH_3OH

3 0

3 years ago

A chemistry student weighs out 0.027 kg of an unknown solid compound X and adds it to 550. mL of distilled water at 30.° C. Afte

QveST [7]

Answer:

1. Yes

2.The solubility of X is 34.55g/L

Explanation:

Solubility of solute refers to how readily a solute will dissolve in a solvent at a particular temperature. Its the amount of moles or grams required to saturate 1dm $^{3}$ or 1 Litre of water.

From the problem, when the liquid was drained off and amount of X which didn't dissolve was measured, it weighed 0.008kg, this means out of 0.027kg, 0.027-0.008 actually dissolved

= 0.019kg*1000 = 19g.

if 19g is required to saturate 550mL at 30°C,

then $\frac{19*1000}{550}$ will saturate 1L

= 34.545g will saturate 1Litre

The solubility thus is 34.55g/L

5 0

3 years ago