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3 years ago

14

A student has a rectangular block. It is 2 cm wide. 3 cm tall and 25cm long. It has a mass of 600 g. First, calculate the volume

of the block:

1 answer:

Klio2033 [76]3 years ago

8 0

Answer:

The area of the block is 150 cm

Explanation:

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How many grams of ice at -13 deg C must be added to 711 grams of water that is initially at a temperature of 87 deg C to produce

Basile [38]

Answer:

The answer to your question is m = 4.7 kg

Explanation:

Data

Ice Water

mass = ? mass = 711 g

T₁ = -13°C T₁ = 87°C

T₂ = 10°C T₂ = 10°C

Ch = 2090 J/kg°K Cw = 4180 J/kg°K

Process

1.- Convert temperature to kelvin

T₁ = 273 + (-13) = 260°K

T₁ water = 87 + 273 = 360 °K

T₂ = 10 + 273 = 283°K

2.- Write the equation of interchange of heat

- Heat lost = Heat absorbed

- mwCw(T₂ - T₁) = miCi(T₂ - T₁)

-Substitution

- 0.711(4180)(10 - 87) = m(2090)(10 - (-13))

- Simplification

228842.46 = 48070m

m = 228842.46/48070

-Result

m = 4.7 kg

4 0

3 years ago

An organic compound is 61.5% C, 2.56% H and 35.9% N by mass. 2.00 grams of this gas is entered into a 300.0 mL flask and heated

pashok25 [27]

Answer:

C₄H₂N₂

Explanation:

First we<u> calculate the moles of the gas</u>, using PV=nRT:

P = 2670 torr ⇒ 2670/760 = 3.51 atm

V = 300 mL ⇒ 300/1000 = 0.3 L

T = 228 °C ⇒ 228 + 273.16 = 501.16 K

3.51 atm * 0.3 L = n * 0.082atm·L·mol⁻¹·K⁻¹ * 501.16 K

n = 0.0256 mol

Now we<u> calculate the molar mass of the compound</u>:

2.00 g / 0.0256 mol = 78 g/mol

Finally we use the percentages given to<em> </em><u>calculate the empirical formula</u>:

C ⇒ 78 g/mol * 61.5/100 ÷ 12g/mol = 4

H ⇒ 78 g/mol * 2.56/100 ÷ 1g/mol = 2

N ⇒ 78 g/mol * 35.9/100 ÷ 14g/mol = 2

So the empirical formula is C₄H₂N₂

6 0

3 years ago

When the following aqueous solutions are mixed together, a precipitate forms. Balance the net ionic equation in standard form fo

rewona [7]

<u>Answer:</u>

<u>For (a):</u> The balanced net ionic equation is $2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$ and the sum of coefficients is 4

<u>For (b):</u> The balanced net ionic equation is $Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$ and the sum of coefficients is 4

<u>For (c):</u> The balanced net ionic equation is $Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$ and the sum of coefficients is

<u>For (d):</u> The balanced net ionic equation is $Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$ and the sum of coefficients is 4

<u>For (e):</u> The balanced net ionic equation is $Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$ and the sum of coefficients is 3

<u>Explanation:</u>

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

For (a): Sodium sulfide and silver nitrate

The balanced molecular equation is:

$Na_2S(aq)+2AgNO_3(aq)\rightarrow 2NaNO_3(aq)+Ag_2S(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+S^{2-}(aq)+2Ag^+(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ag_2S(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (b): Lead(II) nitrate and sodium chloride

The balanced molecular equation is:

$2NaCl(aq)+Pb(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+PbCl_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2Cl^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+PbCl_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (c): Calcium nitrate and potassium carbonate

The balanced molecular equation is:

$K_2CO_3(aq)+Ca(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+CaCO_3(s)$

The complete ionic equation follows:

$2K^{+}(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2K^+(aq)+2NO_3^-(aq)+CaCO_3(s)$

As potassium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

For (d): Barium nitrate and sodium hydroxide

The balanced molecular equation is:

$2NaOH(aq)+Ba(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+Ba(OH)_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2OH^{-}(aq)+Ba^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ba(OH)_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions

The net ionic equation follows:

$Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (e): Silver nitrate and sodium chloride

The balanced molecular equation is:

$NaCl(aq)+AgNO_3(aq)\rightarrow NaNO_3(aq)+AgCl(s)$

The complete ionic equation follows:

$Na^{+}(aq)+Cl^{-}(aq)+Ag^{+}(aq)+NO_3^{-}(aq)\rightarrow Na^+(aq)+NO_3^-(aq)+AgCl(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

8 0

3 years ago

Is it possible to make new water

alexandr1967 [171]

No, Matter cannot be created nor deastroyed.

5 0

3 years ago

Read 2 more answers

What is the partial pressure of C?<br> atm C

s344n2d4d5 [400]

The partial pressure of carbon is 45 mm Hg.

Explanation:

The partial pressure of carbon dioxide is referred as the amount of carbon dioxide present in venous or arterial blood. It acts as a ventilation in the lungs.
There is a formula for measuring partial pressure . As we know total pressure means summation of the pressure of all the gases included .
To find partial pressure we need- total pressure* fraction of mole of that gas. The partial pressure of CO2 is more because it carries deoxygenated blood from the whole body towards the lungs.

8 0

3 years ago