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iren [92.7K]
3 years ago
7

Which of the following methods will usually cause a reaction to speed up? Combining the reactants at a lower temperature. Reduci

ng the concentration of reactants. Grinding one of the reactants into a powder. Removing the catalyst.
Chemistry
2 answers:
Aneli [31]3 years ago
6 0
Ans: grinding one of the reactants into a powder.

this increases the surface area and hence there would be more chances of collisions happening. therefore reaction rate would increase
vredina [299]3 years ago
3 0

The answer is A. Changing the temperature from 90ºC to 40ºC

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33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b
Darina [25.2K]

Explanation:

Mass of fructose = 33.56 g

Mass of water =  18.88  g

Total mass of the solution =  Mass of fructose + Mass of water = M

M = 33.56 g + 18.88  g =52.44 g

Volume of the solution = V = 40.00 mL

Density =\frac{Mass}{Volume}

a) Density of the solution:

\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol

Molar mass of water = 18.02 g/mol

Moles of water= n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol

Mole fraction of fructose in this solution:\chi_1

\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}

\chi_1=0.1510

Mole fraction of water = \chi_2=1-\chi_1=0.8490

c) Average molar mass of of the solution:

=\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol

=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

\frac{\text{Average molar mass}}{\text{Density of the mass}}

=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol

5 0
3 years ago
Write a word equation for the chemical reaction.<br><br> I’ll brainlist.
Alex

Answer:

potassium hydroxide + sulfuric acid → potassium sulfate + water.

Explanation:

Hope it helps.. if yes, plz mark me as brainliest

5 0
3 years ago
The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell
marysya [2.9K]

Answer:

\boxed{\rm \text{Fe(s) $\rightleftharpoons$ Fe$^{2+}$(aq) + 2e$^{-}$}}

Explanation:

The half-cell reduction potentials are

Ag⁺(aq) +   e⁻ ⇌ Ag(s)     E° =  0.7996 V

Fe²⁺(aq) + 2e⁻ ⇌ Fe(s)     E° = -0.447    V

To create a spontaneous voltaic cell, we reverse the half-reaction with the more negative half-cell potential.

The anode is the electrode at which oxidation occurs.

The equation for the oxidation half-reaction is

\boxed{\rm \textbf{Fe(s) $\rightleftharpoons$ Fe$^{2+}$(aq) + 2e$^{-}$}}

4 0
3 years ago
You are given 3.0 grams of solid sodium to react to pure water which has a molarity of 55.6 M. How many milligrams of H2 can be
serious [3.7K]

Answer:

= 15.51 mL

Explanation:

Here's is the reaction:

2HgO(s) ⇒ 2 Hg(s)+O₂(g)

In this reaction 2mol HgO =  1mol O₂

The molecular weight of HgO = 216.59g

so, 3.0g HgO = 3.0g x 1.00molHgO/216.59gHgO

= 0.0138511 molHgO

The amount of Oxygen follows:

0.0138511 molHgOx1/2= 0.00692555 mol O₂

Now, volume of 1 any gas = 22400mL

so, 0.00692555 mol O₂ x22400mLO₂/1mol O₂

= 15.513232mL O₂

4 0
3 years ago
How many moles of glucose can be produced
Arturiano [62]

Answer:

moles of glucose

<u>2.3166 moles of glucose</u>

<u></u>

Explanation:

The balance reaction for the formation of glucose is :

6CO_{2}+6H_{2}O\rightarrow C_{6}H_{12}O_{6}+6O_{2}

here , CO2 = carbon dioxide

H2O = water

C6H12O6 = glucose

O2 = Oxygen

According to this equation :

6 mole of CO2 = 6 mole of H2O = 1 mole of C6H12O6 = 6 mole of O2

We are asked to calculate the mole of Glucose from carbon dioxide.

So,

6 mole of CO2  produce = 1 mole of C6H12O6

1 mole of CO2 will produce =

\frac{1}{6} moles of glucose

13.9 moles of CO2 will produce :

\frac{1}{6}\times 13.9

=2.3166 moles of glucose

Note : first , Always calculate for one mole (By dividing)

. After this , multiply the answer with the moles given.

Always write the substance whose amount is asked(glucose) to the right hand side

5 0
3 years ago
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