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JulijaS [17]
3 years ago
14

Determine the percent composition of sodium in sodium carbonate

Chemistry
1 answer:
crimeas [40]3 years ago
3 0
11×2=22
(11×2)+(6)+(8×3)=52
22/52=0.4230
0.4230×100=42.3%
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What fuel source is Jan using if she exercises at 85% of her maximum aerobic capacity?
Marina86 [1]

Answer:

Carbohydrates

Explanation:

Increased exercise intensity means the overall need for energy increases. As we increase exercise intensity we increase our glucose uptake and oxidation which far exceeds uptake, indicating that muscle stores of glycogen are being used. At moderate intensities (65%) there is an increased need for muscle glycogen and muscle triglycerides which is fat. At higher levels of intensities (85%) there is an even greater need for energy, and this is met almost solely by an increased uptake of glucose from the blood and from muscle glycogen.

In the case of fats as an energy fuel source at high intensities, increasing levels of intensity increases fat oxidation but once we get into higher levels of intensity, we return to levels of fat oxidation similar to very low intensities.

4 0
3 years ago
Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT
marysya [2.9K]

Answer:

(a) The length at temperature 180°C is 40.070 cm

(b) The length at temperature 90°C is 64.976 inches

(c) L(T, α) = 60·α·T - 9000·α + 60

Explanation:

(a) The given parameters are

The thermal expansion coefficient, α for steel = 1.24 × 10⁻⁵/°C

The initial length of the steel L₀ = 40 cm

The initial temperature, t₀ = 40°C

The length at temperature 180°C = L

Therefore, from the given relation, for change in length, ΔL, we have;

ΔL = α × L₀ × ΔT

The amount the temperature changed ΔT = 180°C - 40°C = 140°C

Therefore, the change in length, ΔL, is found as follows;

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 40 × 140°C = 0.07 cm

Therefore, L =  L₀ + ΔL = 40 + 0.07 = 40.07 cm

The length at temperature 180°C = 40.07 cm

(b) Given that the length at T = 120°C is 65 in., we have;

The temperature at which the new length is sought = 90°C

The amount the temperature changed ΔT = 90°C - 120°C = -30°C

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 65 × -30°C = -0.024375 inches

The length, L at 90°C is therefore, L = L₀ + ΔL = 65 - 0.024375 = 64.976 in.

The length at temperature 90°C = 64.976 inches

(c) L = L₀ + ΔL  = L₀ +  α × L₀ × ΔT = L₀ +  α × L₀ × (T - T₀)

Therefore;

L = 60 +  α × 60 × (T - 150°C)

L = 60 + α × 60 × T - 9000 × α

L(T, α) = 60·α·T - 9000·α + 60

7 0
3 years ago
Where are the alkali metals and alkaline earth metals and halogens and noble gases located in the periodic table?
mash [69]
The alkali metals are first column, alkali earth 2nd, halogens 2nd to last, and noble gases last. Hope it helps!
7 0
3 years ago
I have no idea how to do this help plz!
Wittaler [7]
The answes are
C)
Then it is
)A
Im doing this for class too bruh
8 0
3 years ago
Read 2 more answers
As the mass of a star increases the lifespan of the star
ira [324]

Answer:

decreases

Explanation:

4 0
3 years ago
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