17.The column that an element is located in tells us

2CuS + 3O2 = 2CuO + 2SO2

Nata [24]

I have no yuuuuu to say anything abt to

4 0

3 years ago

Define adhesion and give an example?

bulgar [2K]

Similar to Cohesion, adhesion is the attraction between molecules of different substances . Water uses adhesion when it's attraction to another substance is greater than the water's attraction to itself. if you have ever dropped a cup of water on a hardwood floor,you know that it spreads out instead of foaming beads.

3 0

2 years ago

How many grams of iodine are needed to prepare 28.6 grams of ICl

Maslowich

The answer will be 17.9 grams

4 0

3 years ago

A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci

Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

$C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}$

The ICE table is then shown as:

$C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}$

Initial (M) 1.8 0 0

Change (M) - x + x + x

Equilibrium (M) (1.8 -x) x x

$K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}$

where ;

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}$

Since the value for $K_a$ is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

$3.02*10^{-5} *(1.8) = {(x)(x)}$

$5.436*10^{-5}= {(x^2)$

$x = \sqrt{5.436*10^{-5}}$

$x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M$

Dissociated form of 4-chlorobutanoic acid = $C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M$

Percentage dissociated = $\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100$

Percentage dissociated = $\frac{7.3729*10^{-3}}{1.8 }*100$

Percentage dissociated = 0.4096

Percentage dissociated = 0.41% (to two significant digits)

3 0

3 years ago

Read 2 more answers

Plants undergo photosynthesis to produce glucose according to the reaction below. What mass of water is required to produce 5.0g

solniwko [45]

Answer:

option a) 3 g

Explanation:

mass of Glucose = 5 g

Mass of H₂O = ?

Reaction Given:

6CO₂ + 6H₂O ----> C₆H₁₂O₆ + 6O₂

Solution:

First we have to find mass of glucose from balanced reaction.

So,

Look at the reaction

6CO₂ + 6H₂O -------> C₆H₁₂O₆ + 6O₂

6 mol 1 mol

As 6 mole of water (H₂O) give 1 mole of Glucose (C₆H₁₂O₆ )

Convert moles to mass

molar mass of C₆H₁₂O₆ = 6(12) + 12(1) + 6(16)

molar mass of C₆H₁₂O₆ = 72 + 12 + 96

molar mass of C₆H₁₂O₆= 180 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now

6CO₂ + 6H₂O ---------> C₆H₁₂O₆ + 6O₂

6 mol (18 g/mol) 1 mol (180 g/mol)

108 g 180 g

108 g of water (H₂O) produce 180 g of glucose (C₆H₁₂O₆)

So

if 108 g of water (H₂O) produce 180 g of glucose (C₆H₁₂O₆) so how many grams of water (H₂O) will be required to produce 5 g of glucose (C₆H₁₂O₆).

Apply Unity Formula

108 g of water (H₂O) ≅ 180 g of glucose (C₆H₁₂O₆)

X g of water (H₂O) ≅ 5 g of glucose (C₆H₁₂O₆)

Do cross multiply

mass of water (H₂O) = 108 g x 5 g / 180 g

mass of water (H₂O) = 3 g

So 3 g of water is required to produce 5 g of glucose.

7 0

3 years ago