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.)
I assume what you're asking about is, how does the temperature changes when we increase water's mass, according the formula for heat ?
Well the formula is :

(where Q is heat, m is mass, c is specific heat and

is change in temperature. So according this formula, increasing mass will increase the substance's heat, but won't effect it's temperature since they are not related. Unless, if you want to keep the substance's heat constant, in that case when you increase it's mass you will have to decrease the temperature
Answer:

Explanation:
Since the <em>rate constant</em> has units of <em>s⁻¹</em>, you can tell that the order of the reaction is 1.
Hence, the rate law is:
![r=d[A]/dt=-k[A]](https://tex.z-dn.net/?f=r%3Dd%5BA%5D%2Fdt%3D-k%5BA%5D)
Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:
![[A]=[A]_0e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_0e%5E%7B-kt%7D)
You know [A]₀, k, and t, thus you can calculate [A].
![[A]=0.548M\times e^{-3.6\cdot 10^{-4}/s\times99.2s}](https://tex.z-dn.net/?f=%5BA%5D%3D0.548M%5Ctimes%20e%5E%7B-3.6%5Ccdot%2010%5E%7B-4%7D%2Fs%5Ctimes99.2s%7D)
![[A]=0.529M](https://tex.z-dn.net/?f=%5BA%5D%3D0.529M)
A. density
<span>Reaction rate increases with concentration.</span>
Answer:
see explanation...
Explanation:
Mg⁺²-24 Co⁺³-60 Clˉ-35
Protons (p⁺) 12 27 17
Neutrons (n⁰) 12 33 18
Electrons (eˉ) 10 24 18
(c) (b) (a)
12/2 : 12/2 : 10/2 27/3 : 33/3 : 24/3 #n⁰ = 18
6 : 6 : 5 9 : 11 : 8 #eˉ = 18