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3 years ago

PLEASE HELP WITH PAGE 1 FOR QUESTIONS 1, 2, 3 WITH THE GRAPHING THANK YOU.

Chemistry

1 answer:

Scrat [10]3 years ago

6 0

I need a picture in order to answer that question.

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The thin outer layer of Earth, called the crust, contains only 0.50 percent of Earth's total mass and yet is the source of almos

Rzqust [24]

Answer:

The mass of silicon in kilograms in Earth's crust is $7.2729\times 10^{21} kg$ .

Explanation:

Mass of Earth = $5.9\times 10^{21} tons$

(1 ton= 2000 lb)

(1 lb =453.6 g)

1 ton = 2000 × 453.6 g =907,200 g

Mass of Earth = $5.9\times 10^{21}\times 907,200 g=5.3524\times 10^{27} g$

Percentage of earth crust = 0.50%

Mass of earth crust = M

$0.50\%=\frac{M}{5.3524\times 10^{27}}\times 100$

$M=2.6762\times 10^{25} g$

Percentage of the silicon in Earth's crust = 27.2 %

Mass of silicon in in Earth's crust = m

$Si\%=\frac{m}{M}\times 100$

$27.2\%=\frac{m}{2.6762\times 10^{25} g}\times 100$

$m = 7.2792\times 10^{24} g=7.2729\times 10^{21} kg$

1000 g = 1 kg

The mass of silicon in kilograms in Earth's crust is $7.2729\times 10^{21} kg$ .

7 0

3 years ago

How many Liters of space will a 70.0g sample of CO2 occupy?

Tanzania [10]

Answer:

35.6 liters at STP

Explanation:

The molar mass of carbon dioxide is about 44.01 g/mol. The volume of a mole of ideal gas at STP is 22.4 L, so the volume of 70.0 g will be ...

(70.0g)/(44.01 g/mol)·(22.4 L/mol) ≈ 35.6 L

5 0

3 years ago

A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a

Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

$H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)$

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = $m C .$ Δ $T$

m is mass of the solution = 151.8 mL * $\frac{1 g}{1 mL} = 151.8 g$

C is the specific heat of solution = 4.18 $\frac{J}{g. ^{0}C}$

ΔT is the temperature change = $31.50^{0}C - 21.45^{0}C = 10.05^{0}C$

q = $151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J$

Moles of NaOH = $101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol$ NaOH

Moles of $H_{2}SO_{4}$ = $50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}$

Enthalpy of the reaction = $\frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol$

5 0

3 years ago

Which of the following can explain the daily change in sea level observed along a coast?

salantis [7]

The answer for the question above is A. the gravitational pull of the moon on the water near the coast. The sun and and the moon are responsible for the rising and falling of the ocean tides. The gravitational pull of the moon and the sun makes the water in the oceans bulge, causing a continuous change between high and low tide.

5 0

3 years ago

Please help me with these 3 questions with will be giving out brainiest

Anna71 [15]

Answer: All of them are true

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3 years ago