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3 years ago

How do greenhouse gases affect the heat flow into and out of Earth's atmosphere?

1 answer:

3 0

they allow solar energy to penerate the atmosphere and get absorbdd by thr earths surface thus warming it up. they absorb thr out going earths energy keeping thr energy warmth near the earths surface

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Express the concentration of a 0.0320 M aqueous solution of fluoride, F−, in mass percentage and in parts per million (ppm). Ass

denis23 [38]

Answer:

607 ppm

Explanation:

In this case we can start with the ppm formula:

$ppm=\frac{mg~of~solute}{Litters~of~solution}$

If we have a solution of 0.0320 M, we can say that in 1 L we have 0.032 mol of $F^-$ , because the molarity formula is:

$M=\frac{mol}{L}$

In other words:

$0.0320~M=\frac{mol}{1~L}$

$mol=0.032~M*1~L=0.032~mol$

$1~L~of~Solution=0.0320~mol~of~solute$

If we use the atomic mass of $F$ (19 g/mol) we can convert from mol to g:

$0.0320~mol~F^-\frac{19~g~F^-}{1~mol~F^-}~=~0.607~g$

Now we can convert from g to mg (1 g= 1000 mg), so:

$0.607~g\frac{1000~mg}{1~g}=607~mg$

Finally we can divide by 1 L to find the ppm:

$ppm=\frac{607~mg}{1~L}=~607~ppm$

We will have a concentration of 607 ppm.

I hope it helps!

4 0

3 years ago

A 20.0-milliliter sample of 0.200 M K2CO3 solution is added to 30.0 milliliters of 0.400 M Ba(NO3)2 solution.

jenyasd209 [6]

Answer:

(B) 0.160 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $K_2CO_3$ :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of $K_2CO_3$ :

$Moles=0.200 \times {20.0\times 10^{-3}}\ moles$

Moles of $K_2CO_3$ = 0.004 moles

For $Ba(NO_3)_2$ :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

Thus, moles of $Ba(NO_3)_2$ :

$Moles=0.400\times {30.0\times 10^{-3}}\ moles$

Moles of $Ba(NO_3)_2$ = 0.012 moles

According to the given reaction:

$K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}$

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate = 0.004 moles

Available moles of barium nitrate = 0.012 moles

Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

Also,

0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

Also, consumed barium nitrate = 0.004 moles (barium ions precipitate with carbonate ions)

Left over moles = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L

So, Concentration = 0.008/0.05 M = 0.160 M

(B) is correct.

4 0

3 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago

What is the empirical formula for a compound if a sample contains 3.72 g of P and 21.28 g of Cl

fenix001 [56]

M(P)=3.72 g
M(P)=31 g/mol

m(Cl)=21.28 g
M(Cl)=35.5 g/mol

n(P)=m(P)/M(P)
n(P)=3.72/31=0.12 mol

n(Cl)=m(Cl)/M(Cl)
n(Cl)=21.28/35.5=0.60 mol

P : Cl = 0.12 : 0.60 = 1 : 5

PCl₅ - is the empirical formula

6 0

3 years ago

Which answer below correctly the type of change and explanation for the separation of the iron with sulfur?

valina [46]

Chemical reaction is a process in which one set of chemical substances (reactants) is converted into another (products). It involves making and breaking chemical bonds and the rearrangement of atoms. Chemical reactions are represented by balanced chemical equations, with chemical formulas symbolizing reactants and products. For specific chemical reactants, two questions may be posed about a possible chemical reaction. First, will a reaction occur? Second, what are the possible products if a reaction occurs? This

3 0

3 years ago