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NikAS [45]
3 years ago
10

No pic/link pls.

Chemistry
1 answer:
Alex73 [517]3 years ago
7 0
There is no solution but it’s 180.18g
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Which statement is true of a heterogeneous mixture but not of a
Natalka [10]

Answer:

B

Explanation:

well heterogenous mixtures are not uniformly distributed these meaning whatever components are "mixed" can be seperated easily

a example of this can be

Soil, oil in water, ice in water

while Homogeneous mixture cannot be seen seperated

4 0
3 years ago
if a 500.0-g block of metal absorbs 5,016J of heat when its temperature changes from 20.00 Celcius, what is the specific heat?
ZanzabumX [31]
Correction: The temperature change is from 20 °C to 30 °C.

Answer:
               Cp  =  1.0032 J.g⁻¹.°C⁻¹

Solution:

The equation used for this problem is as follow,

                                                  Q  =  m Cp ΔT   ----- (1)

Where;
            Q  =  Heat  =  5016 J

            m  =  mass  =  500 g

            Cp  =  Specific Heat Capacity  =  ??

            ΔT  =  Change in Temperature  =  30 °C  -  20 °C  =  10 °C

Solving eq. 1 for Cp,

                                 Cp  =  Q / m ΔT

Putting values,
                                 Cp  =  5016 J / (500 g × 10 °C)

                                 Cp  =  1.0032 J.g⁻¹.°C⁻¹
4 0
4 years ago
Which will not appear in the equilibrium constant expression for the reaction below?
alexira [117]

a

is the answer i just took the tes

8 0
4 years ago
Read 2 more answers
[24 points] A sample of soil has a total volume of 205 cm3. The soil mass when saturated is 361 g. A specific yield test was con
telo118 [61]

Answer:

Follows are the solution to the given question:

Explanation:

Dry Soil weight = solid soil weight = 284 \ grams

solid soil volume =205 \ cc

saturated mass soil = 361 \ g

The weight of the soil after drainage is =295 \ g

Water weight for soil saturation = (361-284) = 77 \ g

Water volume required for soil saturation =\frac{77}{1} = 77 \ cc

Sample volume of water: = \frac{\text{water density}}{\text{water density input}}

= 361- 295 \\\\ = 66 \ cc

Soil water retained volume = (draining field weight - dry soil weight)

                                             = 295 - 284 \\\\ = 11 \ cc.

\text{POROSITY}= \frac{\text{Vehicle volume}}{\text{total volume Soil}}

                    = \frac{77}{(205 + 77)}  \\\\= \frac{77}{(282)}  \\\\ = 27.30 \%

(Its saturated water volume is equal to the volume of voids)

\text{YIELD SPECIFIC} = \frac{\text{Soil water volume}}{\text{Soil volume total}}

                              = \frac{66}{(205+77)}\\\\= \frac{66}{(282)}\\\\=0.2340\\\\ = 0.23

\text{Specific Retention}= \frac{\text{Volume of soil water}}{\text{Total soil volume}}

                            = \frac{11}{282} \\\\= 0.0390 \\\\ = 0.04

6 0
3 years ago
Why does such a small decrease in pH mean such a large increase in acidity?
den301095 [7]

Answer:

pH is an index of how many protons, or hydrogen ions (H+) are dissolved and free in a solution. The pH scale goes from 0 to 14. A fluid with a pH of 7 is neutral. Below 7, it is acidic; above 7, it is alkaline.

The more below or above 7 a solution is, the more acidic or alkaline it is. The scale is not linear—a drop from pH 8.2 to 8.1 indicates a 30 percent increase in acidity, or concentration of hydrogen ions; a drop from 8.1 to 7.9 indicates a 150 percent increase in acidity. Bottom line: Small-sounding changes in ocean pH are actually quite large and definitely in the direction of becoming less alkaline, which is the same as becoming more acidic.

If you think about it, we use descriptive words like this all the time. A person who stands 5’5” tall and weighs 300 pounds isn’t thin. If he loses 100 pounds, he still won’t be thin, but he will be thinner than he was before he went on the diet. (And we are more likely to comment that he’s looking trimmer than to say he’s not as fat as he used to be.)

7 0
3 years ago
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