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4 years ago

9

What type of energy can be moved through conductors?

2 answers:

LenaWriter [7]4 years ago

6 0

Light and energey!!!!

laiz [17]4 years ago

3 0

Electric energy that should be the answer

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What is the mass of volume= 5.4 mL density= 2.5 g/mL

ddd [48]

Answer:

13.5g

Explanation:

Mass is defined as the measure of the amount of matter in an object. Its unit is kg or g.

Mass can be calculated using the formula:

m= d × v

where,

d= density

m= mass

v= volume

m= 2.5×5.4

m= 13.5g

5 0

3 years ago

Give one chemical property of soda ash.

Elenna [48]

<span>Soda ash is sodium carbonate, Na2CO3. One chemical property of this compound is its basicity, which is measured by the pKb. The pKb for sodium carbonate is 3.67. It is the result of the dissociation of Na2CO3 in water: Na2CO3 + H2O = Na HCO3 + Na (+) + OH(-). This pKb means that it is a highly basic compound. pKb = log { 1 / [OH-] }, so pKb is a measure of the concentrations of OH- ions, which is the basiciity of the compound. </span>

4 0

3 years ago

Protons have a positive charge true or false

aksik [14]

Answer:

True

Explanation:

Protons have a positive electrical of +1

6 0

3 years ago

Help if possible!<br> Picture attached with the problems.

MAVERICK [17]

1. 1 M , 2 M , 1 M
2. 10 mol , 0.1 mol , 0.5 mol
3. 0.5 L , 6.6 L , 5/21 L

M = mol/L

8 0

3 years ago

The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are present in 0.50 l of a soluti

lyudmila [28]

$NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-}$

Here the base is a benzoate ion, which is a weak base and reacts with water.

$C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq)$

The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.

Therefore [OH-] = [C6H5COOH]

In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]

pOH = 14 - pH

pH given = 9.04

pOH = 14-9.04 = 4.96

pOH = -log[OH-] or $[OH^{-}] = 10^{^{-pOH}}$

$[OH^{-}] = 10^{^{-4.96}}$

$[OH^{-}] = 1.1\times 10^{-5}$

The base dissociation equation kb = $\frac{Product}{Reactant}$

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.

Value of Kb is given = $1.6\times 10^{-10}$

And value of [OH-] we have calculated as $1.1\times 10^{-5}$ and value of C6H5COOH is equal to OH-

Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

$1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}$

$[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}$

$[C6H5COO^{-}] = 0.76 M$ or $0.76\frac{mol}{L}$

So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L

Moles of NaC6H5COO would be = $0.76(\frac{mol}{L}) \times (0.50L)$

Moles of NaC6H5COO (sodium benzoate) = 0.38 mol

8 0

3 years ago

Read 2 more answers