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Nutka1998 [239]

2 years ago

7

a classroom has an aquarium. In the aquarium are a goldfish and a water plant. the teacher explains that both organisms use cell

ular respiration to transform food into energy. identify how the plant and the goldfish get the materials needed for cellular respiration.

1 answer:

Anna71 [15]2 years ago

4 0

the thing that helps is always gonna help that’s why u should always help

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Plz answer thx<br><br><br> How many moles are in 2.5 x 10^18 grams of water?

alex41 [277]

Answer:

$1.39\times 10^{17}$ moles of water in $2.5\times 10^{18} g$ of water.

Explanation:

Mass of water = $m = 2.5\times 10^{18} g$

Molar mass of water = M = 18 g/mol

Moles = n = $\frac{m}{M}$

$n=\frac{2.5\times 10^{18} g}{18 g/mol}=1.39\times 10^{17} moles$

So, there are $1.39\times 10^{17}$ moles of water in $2.5\times 10^{18} g$ of water.

3 0

3 years ago

Ayudaaaaaa!!!!!!!!!!!!!!!!!!

Eva8 [605]

Answer:

uh you good

Explanation:

4 0

2 years ago

A t-test is suitable for

blagie [28]

Answer:

Explanation:

A t-test is a type of inferential statistic used to determine if there is a significant difference between the means of two groups, which may be related in certain features. The t-test is one of many tests used for the purpose of hypothesis testing in statistics. Calculating a t-test requires three key data values.

7 0

3 years ago

Is a cube of salt being crushed before being stirred into water a physical or chemical change

Olin [163]

By crushing the salt, you are performing a physical change because you aren't altering the chemical makeup of the salt, just the physical form. Hope this helps! :)

6 0

2 years ago

Read 2 more answers

in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc

Margarita [4]

Answer: 0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$ .....(1)

a) Molarity of $Na_2S$ solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

$0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol$

$\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol$

b) Molarity of $Hg(NO_3)_2$ solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

$0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol$

$Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3$

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ reacts with 1 mole of $Na_2S$

Thus 0.0005 moles of $HgNO_3$ reacts with= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg(NO_3)_2$

Thus $Hg(NO_3)_2$ is the limiting reagent and $Na_2S$ is the excess reagent.

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ forms= 1 mole of $Hg_2S$

Thus 0.0005 moles of $Hg(NO_3)_2$ forms= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg_2S$

mass of $H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g$

Thus 0.1161 grams of mercury(II) sulfide) form.

5 0

3 years ago