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3 years ago

10

In a reaction that occurs in solution, the volume will not change. What happens to the concentration of the reactants? What happ

ens to the reaction rate as the reaction proceeds?
(will give brainllest)

1 answer:

Stella [2.4K]3 years ago

6 0

Answer:

the reaction will increase

Explanation:

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Addie mixes together a solution of 10 grams of

Gnesinka [82]

Answer: 150 grams

Explanation: m = V × ρ

= 15 milliliter × 10 gram/cubic centimeter

= 15 cubic centimeter × 10 gram/cubic centimeter

= 150 gram

5 0

3 years ago

SOMEONE PLS HELP

Gekata [30.6K]

Answer:

<h2>464.85 mL</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we're finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

100.7 kPa = 100,700 Pa

95.1 kPa = 95,100 Pa

We have

$V_2 = \frac{100700 \times 439}{95100} = \frac{44207300}{95100} \\ = 464.8506...$

We have the final answer as

<h3>464.85 mL</h3>

Hope this helps you

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3 years ago

Two glucose molecules can combine to form a disaccharide molecule and what molecule

elena-14-01-66 [18.8K]

And a water molecule, this is called a dehydration synthesis. when 2 molecule combine, a water molecule leave.

5 0

3 years ago

8. Calculate the number of moles of eachsubstance.a. 5.45 x 1026 particles of methane, CH4

Klio2033 [76]

<em>ANSWER</em>

The number of moles of methane is 905.32 moles

STEP-BY-STEP EXPLANATION:

Given information

The number of particles of methane = 5.45 x 10^26 particles

Let x represents the number of moles of methane

To calculate the number of moles, we will be using the below formula

$\text{Number of particles = number of moles x Avogadro's constant}$

Recall that, the Avogadro's constant is given as

$6.02\cdot10^{23}$ $\begin{gathered} 5.45\cdot10^{26}\text{ = x }\cdot\text{ 6.02 }\cdot10^{23} \\ \text{Divide both sides by 6.02 }\cdot10^{23} \\ x\text{ = }\frac{5.45\cdot10^{26}}{6.02\cdot10^{23}} \\ x\text{ = }\frac{5.45}{6.02}\cdot10^{26\text{ - 23}} \\ x\text{ = 0.9053 }\cdot10^3 \\ x\text{ = 905.32 moles} \end{gathered}$

Therefore, the number of moles of methane is 905.32 moles

6 0

1 year ago

How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?

meriva

Answer:

$\large \boxed{\text{77.4 mL}}$

Explanation:

Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

V/mL: 249

c/mol·L⁻¹: 0.0443 0.285

1. Calculate the moles of Ba(OH)₂

$\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}$

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

$\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}$

3. Calculate the volume of HCl

$V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}$

8 0

3 years ago