Answer is c photosynthesis
Answer:
6. 7870 kg/m³ (3 s.f.)
7. 33.4 g (3 s.f.)
8. 12600 kg/m³ (3 s.f.)
Explanation:
6. The SI unit for density is kg/m³. Thus convert the mass to Kg and volume to m³ first.
1 kg= 1000g
1m³= 1 ×10⁶ cm³
Mass of iron bar
= 64.2g
= 64.2 ÷1000 kg
= 0.0642 kg
Volume of iron bar
= 8.16 cm³
= 8.16 ÷ 10⁶


Density of iron bar

= 7870 kg/m³ (3 s.f.)
7.

Mass
= 1.16 ×28.8
= 33.408 g
= 33.4 g (3 s.f.)
8. Volume of brick
= 12 cm³

Mass of brick
= 151 g
= 151 ÷ 1000 kg
= 0.151 kg
Density of brick
= mass ÷ volume

(3 s.f.)
Answer:
449 (g K) / 39.1 (g/mol K) = 11.5 mol K
Explanation:
Potassium has atomic number 39.1
amount of K in 449g sample = 449/39.1 = 11.5 mol
Answer:
Explanation:
(NH4)3 PO4 +NaOH arrow Na3PO4 +3NH3 +3H2O
Start by seeing what happens with the Na. You need 3 on the left, so put a 3 in front of NaOH
(NH4)3 PO4 +3NaOH arrow Na3PO4 +3NH3 +3H2O Next work with the nitrogens. YOu have 3 on the left and 3 on the right, so they are OK. Next Go to the stray oxygens.
You have 3 on left in (NaOH) and three on the right in 3H2O so they are fine as well. The last thing you should look at are hydrogens.
There are 12 + 3 on the left which is 15. There are 9 (in 3NH3) and 6 more in the water. They seem fine.
Why didn't I do something with the PO4^(-3)? The reason is a deliberately stayed away from them and balanced everything else. Since they were untouched with 1 on the left and 1 on the right, they are balanced.
Species Na H O N PO4
Left 3 15 3 3 1
Right 3 15 3 3 1
Answer:
The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>
Explanation:
Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M
Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M
<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)
The equilibrium constant (
) for the given chemical reaction, is given by the equation:
![K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%20%7B%5BHI%5D%5E%7B2%7D%7D%7B%5BH_%7B2%7D%5D%5C%3A%20%5BI_%7B2%7D%5D%7D)
<u><em>At the original equilibrium state:</em></u>

<u><em>Therefore, at the new equilibrium state:</em></u>
![\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88](https://tex.z-dn.net/?f=%5CRightarrow%20%5BHI%5D%5E%7B2%7D%20%3D%20713.59%20%5Ctimes%200.01805%20%3D%2012.88)
![\Rightarrow [HI] = \sqrt {12.88} = 3.589 M](https://tex.z-dn.net/?f=%5CRightarrow%20%5BHI%5D%20%3D%20%5Csqrt%20%7B12.88%7D%20%3D%203.589%20M)
<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>