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3 years ago

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What substances were found in the innermost regions (within about the inner 0.3 au) of the solar system before planets began to

form? view available hint(s) what substances were found in the innermost regions (within about the inner 0.3 ) of the solar system before planets began to form? nothing at all only rocks and metals only hydrogen compounds only hydrogen and helium gases rocks, metals, hydrogen compounds, hydrogen, and helium, all in gaseous form?

1 answer:

coldgirl [10]3 years ago

5 0

The answer would be Rocks, metals, hydrogen compounds, hydrogen and helium, all in gaseous form.

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When entering coefficients to balance equations, recall that just writing the symbol implies "1." Therefore, if a coefficient of

ANEK [815]

The balanced equation is MgCl₂ +2AgNO₃ ⟶ 2AgCl + Mg(NO₃)₂

<em>Step 1</em>. Start with the most complicated-looking formula [Mg(NO₃)₂].

Put a 1 in front of it and balance its atoms.

MgCl₂ +AgNO₃ ⟶ AgCl + <u>1</u>Mg(NO₃)₂

<em>Step 2</em>. Balance Mg.

<u>1</u>MgCl₂ +AgNO₃ ⟶ AgCl + <u>1</u>Mg(NO₃)₂

Already balanced —1 atom each side.

<em>Step 3</em>. Balance N.

We have 2 N on the right-hand side and 1 N on the left. We need 2 N on the left. Put a 2 in front of AgNO₃.

<u>1</u>MgCl₂ +<u>2</u>AgNO₃ ⟶ AgCl + <u>1</u>Mg(NO₃)₂

<em>Step 4</em>: Balance O

Done. We have 6 O on the left and 6 O on the right.

<em>Step 5</em>. Balance Ag.

We have 2 Ag on the left, so we need 2 Ag on the right. Put a 2 in front of AgCl.

<u>1</u>MgCl₂ +<u>2</u>AgNO₃ ⟶ <u>2</u>AgCl + 1Mg(NO₃)₂

The balanced equation is

MgCl₂ +2AgNO₃ ⟶ 2AgCl + Mg(NO₃)₂

6 0

4 years ago

On a warm summer day in Oslo, the temperature is 21 degree C and the relative humidity is 70%. What is the absolute humidity? Wh

lesya [120]

Explanation:

According to the psychrometric chart at dry bulb temperature of $21^{o}C$ and RH 70%, the absolute humidity = 0.011 kg/kg dry air

Formula to calculate humid volume is as follows.

$\nu_{H} m^{3}/kg$ dry air = $\frac{22.41}{273}TK (\frac{1}{28.97} + \frac{1}{18.02}H)$

= $(2.83 \times 10^{-3} + 4.56 \times 10^{-3} H) T K$

= $(2.83 \times 10^{-3} + 4.56 \times 10^{-3} \times 0.011) \times 294$

= 0.847 $m^{3}/kg$ dry air

Hence, humid volume of air is 0.847 $m^{3}/kg$ dry air.

Specific enthalpy of dry air = specific heat capacity of dry air × dry bulb temperature

= $1.006 kJ/kg ^{o}C \times 21^{o}C$

= 21.126 kJ/kg dry air

Hence, the specific enthalpy of the air is 21.126 kJ/kg dry air.

As per the psychrometric chart at given conditions wet bulb temperature = 17.5 $^{o}C$

As per the psychrometric chart at given conditions dew point temperature = 15.5 $^{o}C$

4 0

3 years ago

Write conversion factors (as ratios) for the number of:

nignag [31]

The answer is C, grams in 1 ounce, I am pretty sure.

3 0

3 years ago

5. What is a habitat?

svetlana [45]

Answer:

c) The natural home or environment of an animal, plant, or other organism

Explanation:

Please note that in order to post questions in Science, you must post in Science section, but not in History. You are asking a wrong question in the wrong section.

<u><em>Thank You! Please Mark Brainliest!</em></u> :)

6 0

3 years ago

Read 2 more answers

A solution contains two isomers, n-propyl alcohol and isopropyl alcohol, at 25°C. The total vapor pressure is 38.6 torr. What ar

Leona [35]

Answer:

Mole fraction of alcohols in liquid phase $x_1=0.2727\& x_2=0.7273$ .

Mole fraction of alcohols in vapor phase $y_1=0.1468\& y_2=0.8516$ .

Explanation:

The total vapor pressure of the solution = p =38.6 Torr

Partial vapor pressure of the n-propyl alcohol = $p^{o}_1=21.0 Torr$

Partial vapor pressure of the isopropyl alcohol = $p^{o}_2=45.2 Torr$

$p=x_1\times p^{o}_1+x_2\times p^{o}_2$ (Raoult's Law)

$p=x_1\times p^{o}_1+(1-x_1)\times p^{o}_2$

$38.6 Torr=x_1\times 21.0 Torr+(1-x_1)\times 45.2 Torr$

$x_1=0.2727$

$x_2=1-0.2727=0.7273$

$x_1\& x_2$ is mole fraction in liquid phase.

Mole fraction of components in vapor phase $y_1\& y_2$

$p_1=y_1\times p$ (Dalton's law of partial pressure)

$y_1=\frac{p_1}{38.6 Torr}=\frac{p^{o}_1\times x_1}{38.6 Torr}$

$y_1=\frac{21.0 Torr\times 0.2727}{38.6 Torr}=0.1468$

$y_1=\frac{p_2}{38.6 Torr}=\frac{p^{o}_2\times x_2}{38.6 Torr}$

$y_2=\frac{45.2 Torr\times 0.7273}{38.6 Torr}=0.8516$

Mole fraction of alcohols in vapor phase $y_1=0.1468\& y_2=0.8516$

5 0

3 years ago