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3 years ago

How water deposits soil sediment and rock

1 answer:

6 0

Water deposits soil ,sediments ,and rock by moving them to a different place .Say like a beach .The ocean is currently moving the bits of sand and (glass:sand)rock pieces and always ending up in new places.Same as a river.!!!!

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Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t

Rasek [7]

Answer:

$4.6*10^{-14}$ M

Explanation:

Concentration of $Cu^{2+} = [Cu(NO_3)_2]$ = 0.020 M

Constructing an ICE table;we have:

$Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}$

Initial (M) 0.020 0.40 0

Change (M) - x - 4 x x

Equilibrium (M) 0.020 -x 0.40 - 4 x x

Given that: $K_f =1.7*10^{13}$

$K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}$

$1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}$

Since x is so small; 0.40 -4x = 0.40

Then:

$1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}$

$1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}$

$1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x$

$8.704*10^9-4.352*10^{11}x =x$

$8.704*10^9 = 4.352*10^{11}x$

$x = \frac{8.704*10^9}{4.352*10^{11}}$

$x = 0.0199999999999540$

$Cu^{2+}= 0.020 - 0.019999999999954$

$Cu^{2+} = 4.6*10^{-14}$ M

8 0

2 years ago

PLEASE HELPPPP!!!!!

Likurg_2 [28]

Answer : The mass of nitric acid is, 214.234 grams.

Solution : Given,

Moles of nitric acid = 3.4 moles

Molar mass of nitric acid = 63.01 g/mole

Formula used :

$\text{Mass of }HNO_3=\text{Moles of }HNO_3\times \text{Molar mass of }HNO_3$

Now put all the given values in this formula, we get the mass of nitric acid.

$\text{Mass of }HNO_3=(3.4moles)\times (63..01g/mole)=214.234g$

Therefore, the mass of nitric acid is, 214.234 grams.

6 0

2 years ago

If you have 150 grams of s8, how many moles of so2 can you make?

ruslelena [56]

Answer : You can make 4.68 moles of SO₂

Explanation:

Step 1 : Write balanced equation.

S₈ can combine with oxygen to form SO₂ gas. The balanced equation for this reaction is written below.

$S_{8} + 8 O_{2} \rightarrow 8 SO_{2}$

Step 2 : Find moles of S₈

The formula to calculate mol is

$mole = \frac{grams}{MolarMass}$

Molar mass of S₈ is 256.5 g/mol

$mole = \frac{150grams}{256.5 g/mol}$

$mole = 0.585 mol$

we have 0.585 mol of S₈

Step 3: Use mole ratio to find moles of SO₂

The mole ratio of S₈ and SO₂ can be found using balanced equation which is 1:8

That means 1 mol of S₈ can form 8 moles of SO₂.

Let us use this as a conversion factor to find moles of SO₂

$0.585 mol (S_{8}) \times \frac{8 mol (SO_{2})}{1 mol (S_{8})} = 4.68 mol$

Therefore we have 4.68 moles of SO₂

4 0

2 years ago

Read 2 more answers

Calculate the number of moles contained in 550ml of carbon dionide at stp

g100num [7]

The answer is 20 I think

3 0

3 years ago

The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentrat

Valentin [98]

Answer:

(a)

$0.0342M$

(b)

$t_{1/2}=17.36s\\t_{1/2}=23.15s$

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

$\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\$

$[NOBr]=\frac{1}{29.2/M}=0.0342M$

(b) Now, for a second-order reaction, the half-life is computed as shown below:

$t_{1/2}=\frac{1}{k[NOBr]_0}$

Therefore, for the given initial concentrations one obtains:

$t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.072M}=17.36s\\t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.054M}=23.15s$

Best regards.

8 0

2 years ago

Read 2 more answers