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adelina 88 [10]

3 years ago

10

Hana fills a cup with sandy ocean water. She pours the mixture through a filter. What does she collect that passes through the f

ilter? a sample of pure water a solution of salt in water a suspension of sand in water a colloid of salt in water

2 answers:

Neporo4naja [7]3 years ago

8 0

Answer:

a solution of salt in water

Explanation:

On edgenuity 2020

LekaFEV [45]3 years ago

7 0

a solution if salt in water

Hope it helps

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When 4.41g of phosphoric acid (H3PO4) react with 9.25g of barium hydroxide, water and insoluble barium phosphate form. [T/I-7] a

AnnZ [28]

Answer:

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

Explanation:

Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.

H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

We will balance it using the trial and error method.

First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

Finally, we will get the balanced equation by multiplying H₂O by 6.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

3 0

3 years ago

A sucrose solution is prepared to a final concentration of 0.250 M . Convert this value into terms of g/L, molality, and mass %.

MArishka [77]

Answer:

A. 85.6 g

= 0.0856 kg.

B. 0.00027 mol/g

= 0.27 mol/kg.

C. 8.39 %

Explanation:

Given:

Molar concentration = 0.25 M

Molar weight of sucrose = 342.296 g/mol

Density of solution = 1.02 g/mL

Mass of water = 934.4 g.

Density in g/l = 1.020 g/ml * 1000ml/1 l

= 1020 g/l

Mass of solution in 1 l of solution = 1020 g

Mass of solution = mass of solvent + mass of solute

Mass of sucrose = 1020 - 934.4

= 85.6 g of sucrose in 1 l of solution.

A.

Density of sucrose = mass/volume

= molar mass/molar concentration

= 342.296 * 0.25

= 85.6 g/l

Number of moles = mass/molar mass

= 85.6/342.296

= 0.25 mol

B.

Molality = number of moles of solute/mass of solvent

= 0.25/934.4

= 0.00027 mol/g

C.

% mass of sucrose = mass of sucrose/total mass of solution * 100

= 85.6/1020 * 100

= 8.39 %

6 0

2 years ago

(6) Compare a CSTR with a PFR below. a. A flow of 0.3 m3/s enters a CSTR (volume of 200 m3) with an initial concentration of spe

Dmitry [639]

Answer:

Explanation:

Given that:

The flow rate Q = 0.3 m³/s

Volume (V) = 200 m³

Initial concentration $C_o$ = 2.00 ms/l

reaction rate K = 5.09 hr⁻¹

Recall that:

$time (t) = \dfrac{V}{Q}$

$time (t) = \dfrac{200}{0.3}$

$time (t) = 666.66 \ sec$

$time (t) = \dfrac{666.66 }{3600} hrs$

$time (t) = 0.185 hrs$

$\text{Using First Order Reaction:}$

$\dfrac{dc}{dt}=kc$

where;

$t = \dfrac{1}{k} \Big( \dfrac{C_o}{C_e}-1 \Big)$

$0.185 = \dfrac{1}{5.09} \Big ( \dfrac{200}{C_e}- 1 \Big)$

$0.942 = \Big ( \dfrac{200}{C_e}- 1 \Big)$

$1+ 0.942 = \Big ( \dfrac{200}{C_e} \Big)$

$\dfrac{200}{C_e} = 1.942$

$C_e = \dfrac{200}{1.942}$

$\mathbf{C_e = 102.98 \ mg/l}$

Thus; the concentration of species in the reactant = 102.98 mg/l

b). If the plug flow reactor has the same efficiency as CSTR, Then:

$t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3 hrs} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3*3600 sec} =0.196(0.663)$

$V_{PFR} =0.196(0.663)*0.3*3600$

$V_{PFR} = 140.34 \ m^3$

The volume of the PFR is ≅ 140 m³

3 0

3 years ago

lanets formed from ____ left over from the formation of the Sun at the center of the cloud. a. gravel c. solar wind b. gas, ice,

vodka [1.7K]

Might be solar but I may be wrong

4 0

3 years ago

Read 2 more answers

Suppose an egyptian mummy is discovered in which the amount of carbon-14 present is only about oneone-fifthfifth the amount fo

poizon [28]

We are given the following equation:

y = y0 e^-0.0001216 t

where y = 1/5 y0, y0 is the original amount

So solving for time t:

1/5 y0= y0 e^-0.0001216 t

t = 13,235.51 years

So the human died about 13235.5 years ago

3 0

3 years ago