True or false can intrusive igneous rock has larger crystal because it was formed when manga cooled underground?

A compound needed for photosynthesis

Pavel [41]

Answer:

Carbon dioxide as well as water are needed for photosynthesis.

Explanation:

8 0

3 years ago

Select all that apply. choose the two correct statements below. the elements of the halogen family are very reactive because: th

Harman [31]

Halogens are a group of elements consisting of Fluorine, Chlorine, Bromine, Iodine and Astatine. In their ionic form, they have a superscript of -1, for example, chloride ion is Cl-1. These means that they readily accept one electron in order to achieve the Octet rule. The Octet rule states that each atom must contain 8 electrons in their valence shell for it to be stable. The most stable set of elements are the noble gases. Because they already fulfill the Octet rule, they no longer take part in reactions. Halogens are also very electronegative, meaning, they attract more electrons toward them. This is also a consequence of the Octet rule.

From the choices, the answers would be:
<span>they require only one electron to complete their outer shell
they have a high electronegativity</span>

6 0

4 years ago

Please help me! Asap

densk [106]

Correct answer is 20 kJ; 50 kJ

potential energy is resting energy, that peak that's 75 kJ would be the kinetic energy

3 0

4 years ago

The combustion of glucose is represented by the following balanced equation: C6H12O6+6 O2→6 H2O+6 CO2. The reaction uses 1 gram

hichkok12 [17]

Answer : The correct option is, (d) 80 %

Solution : Given,

Mass of $C_6H_{12}O_6$ = 1 g

Mass of $O_2$ = 1 g

Molar mass of $C_6H_{12}O_6$ = 180 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of $C_6H_{12}O_6$ and $O_2$ .

$\text{ Moles of }C_6H_{12}O_6=\frac{\text{ Mass of }C_6H_{12}O_6}{\text{ Molar mass of }C_6H_{12}O_6}=\frac{1g}{180g/mole}=0.00555moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{1g}{32g/mole}=0.0312moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$C_6H_{12}O_6+6O_2\rightarrow 6H_2O+6CO_2$

From the balanced reaction we conclude that

As, 6 mole of $O_2$ react with 1 mole of $C_6H_{12}O_6$

So, 0.0312 moles of $O_2$ react with $\frac{0.0312}{6}=0.0052$ moles of $C_6H_{12}O_6$

From this we conclude that, $C_6H_{12}O_6$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$

From the reaction, we conclude that

As, 6 mole of $O_2$ react to give 6 mole of $H_2O$

So, 0.0312 mole of $O_2$ react to give 0.0312 mole of $H_2O$

Now we have to calculate the mass of $H_2O$

$\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O$