How does the heat content of the reaction change in the process of photosynthesis when a glucose molecule is formed?

What does the pacific tsunami warning system use to detect tsunamis ?

S_A_V [24]

Answer:

C

Explanation:

Their are pressure sensers in the water that will detect high pressure and set of scales that are currently detecting semic waves and trigger sierns.

7 0

2 years ago

Explain how water might have two different densities

MrMuchimi

Answer:

I think this is it

Water can have two different densities if it has substances dissolved in it. ... When liquid water freezes it becomes solid water or ice, which is less dense than liquid water. The fact that solid water (ice) is less dense than liquid water is evident in the way ice floats in a glass of water.

Explanation:

6 0

2 years ago

Two grams of potassium chloride are completely dissolved in a sample of water in a beaker. This solution is classified as

Eddi Din [679]

The answer is (3) a homogeneous mixture. The difference between homogeneous and heterogeneous mixture is the degree of the mixture being mixed. Due to the completely dissolved and the dissolving ability of KCl, we can get the answer,

5 0

3 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago

Convert the following masses into kilograms:

beks73 [17]

Answer:

(a): 2,300 kilograms

(b): 0.005 kilograms

(c): 2.3 × 10^-5 kilograms

(d): 155 kilograms

Explanation:

Formulas:

(a); divide the mass value by 1000

(b); divide the mass value by 1e+6

(c); divide the mass value by 1e+9

(d); multiply the mass value by 1000

5 0

3 years ago