What element would be found in period 5 that has similar chemical properties to Al?

Answer:

Yellow fruit and dwarf vines are recessive traits in tomatoes. Red fruit and tall vines are dominant. Complete a Punnett square and answer the questions for a completely dominant red and tall plant crossed with a heterozygous red and dwarf plant.

Explanation:

Answer:

They are different because one is someone who studies volcanoes and the other is someone who studies weather

Explanation:

Answer:

D. One dependent variable and one experimental variable

Explanation:

The experimental variable is defined as an independent variable which can be manipulated during an experiment in order to find out its impact or influence on the dependent variable.

The dependent variable refers to a variable that changes when an experimental or  independent variable is manipulated.

Every experiment must have at least one experimental variable and one dependent variable in order to yield any meaningful result.

Answer:

Heated water begins circulating in a fish tank.

Explanation:

Answer:

Heated water begins circulating in a fish tank.

Answer:

A.) an exothermic reaction that has an increase in disorder.

Explanation:

got it correct (unit review test)

Answer:

The driver in the car because a truck is a bigger vehicle and the impact of a truck to a car is more damaging than a car to a truck. Weight difference is the key I think.

Explanation:

Solution :

Given :

length of silver cube = 2.28 cm

length of gold cube = 2.75 cm

Initial temperature = 82.8°C

Volume of silver cube is

Volume [tex]$=(\text{edge length})^3$[/tex]

               =    [tex]$(2.28)^3$[/tex]

                [tex]$= 11.8 \ cm^3$[/tex]

mass of silver cube

Mass, [tex]$m_s = \text{density} \times \text{volume} $[/tex]

                = 10.5 x 11.8

                = 123.9 g

Similarly, the volume of gold cube is

Volume [tex]$=(\text{edge length})^3$[/tex]

               =    [tex]$(2.75)^3$[/tex]

                [tex]$= 20.79 \ cm^3$[/tex]

mass of gold cube

Mass, [tex]$m_g = \text{density} \times \text{volume}$[/tex]

                = 19.3 x 20.79

                = 401.247 g

Now

heat lost by silver and gold cube = heat gained by water

∴[tex]$m_g .c_g \Delta T + m_s .c_s \Delta T = m_w.c_w \Delta T$[/tex]

[tex]$m_g .c_g (82.8- T) + m_s .c_s (82.8- T) = m_w.c_w (T - 20.2)$[/tex]

[tex]$401.247 \times 0.1264 (82.8- T) + 123.9 \times 0.2386 (82.8- T) = 111.5 \times 4.184 (T - 20.2)$[/tex]

Now solving the equation

[tex]$50.71 (82.8- T) + 29.56 (82.8- T) = 466.51 (T - 20.2)$[/tex]

Final temperature, T = 31.27°C

Answer:

no .........................

Answer: with dog figs- it is 590

With decimals it is 589.68

Explanation:

Answer:

The molar concentration of water sample that has 1 ppm arsenic is 1.33  * 10⁻⁵ M

Explanation:

1 parts per million (ppm) means 1 part of solute in a million parts of solution.

Since 1 million, 10⁶ mg = 1 kg, 1 ppm = 1 mg/kg

The density of water = 1.0 kg/L, therefore, the volume of 1 kg of water = 1 L

Hence, 1 ppm = 1 mg/kg = 1 mg/L

molar mass of arsenic = 75 g/mol

1 mg of arsenic = 10⁻³ g or 0.001 g

number of moles of arsenic in 0.001 g of arsenic = 0.001 g / 75 g/mol = 1.33  * 10⁻⁵ moles

Molar concentration = 1.33 * 10⁻⁵ mol / 1 L = 1.33  * 10⁻⁵ M

Therefore, the molar concentration of water sample that has 1 ppm arsenic is 1.33  * 10⁻⁵ M

x CO₂  = 0.236

x H₂  = 0.764

Given

P CO₂ = 217 mmHg

P H₂ = 703 mmHg

Required

The mole fraction

Solution

Dalton's law :

The total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases  

Can be formulated:  

P tot = P1 + P2 + P3 ....  

P tot for the mixture :

= 217 mmHg + 703 mmHg

= 920 mmHg

Mole fraction of each gas :

x CO₂ = 217 mmHg/920 mmHg = 0.236

x H₂ = 703 mmHg/920 mmHg = 0.764

Answer:

2 Al(s) + 3 H₂SO₄(aq) ⇒ Al₂(SO₄)₃(aq) + 3 H₂(g)

Explanation:

Let's consider the unbalanced equation for the single replacement reaction between Al and H₂SO₄.

Al(s) + H₂SO₄(aq) ⇒ Al₂(SO₄)₃(aq) + H₂(g)

We will begin balancing S atoms by multiplying H₂SO₄ by 3.

Al(s) + 3 H₂SO₄(aq) ⇒ Al₂(SO₄)₃(aq) + H₂(g)

Then, we balance Al atoms by multiplying Al by 2.

2 Al(s) + 3 H₂SO₄(aq) ⇒ Al₂(SO₄)₃(aq) + H₂(g)

Finally, we get the balanced equation by multiplying H₂ by 3.

2 Al(s) + 3 H₂SO₄(aq) ⇒ Al₂(SO₄)₃(aq) + 3 H₂(g)

Answer:

communicable and non-communicable?

1.63 M

Math

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Chemistry

Unit 0

Aqueous Solutions

Step 1: Define

1.43 mol NaCl (sodium chloride)

0.875 L solution

Step 2: Find M

Step 3: Check

Follow sig fig rules and round. We are given 3 sig figs.

1.63429 M ≈ 1.63 M

Answer:

2KOH + 3co2 ------->2ксоз + 2Н20

Answer:

28

Explanation:

electron = atomic number - charge(if the charge is positive and addition if the charge is negative)

electron = 30 - 2

electron = 28

Explanation:

V is not an SI base unit. (C)

Volume can be expressed as m³ in units.

In turn, m is the SI base units for length.

A scientist exploring evidence from other fields of science to support findings is an example of a scientist using creativity.

Creativity is a mental cognitive ability that involves the creation and/or discovery of a hidden process.

In science, creativity refers to the application of mental skills to create new experimental designs and extrapolate findings.

In conclusion, a scientist exploring evidence from other fields of science to support findings is an example of a scientist using creativity.

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(NH₄)₂S

%N=  41.18%

%H =11.76%

%S=47.06%

Mg(NO₃)₂

%Mg=32.43%

%N=18.92%

%O=(64.86%

Given

(NH₄)₂S

Mg(NO₃)₂

Required

The percent composition

Solution

(NH₄)₂S MW=68 g/mol

%N=2. Ar N  / MW (NH₄)₂S x 100%

%N= (2.14/68) x 100% = 41.18%

%H =(8.1/68) x 100%=11.76%

%S=(32/68)x 100%=47.06%

Mg(NO₃)₂ MW=148 g/mol

%Mg=(2.24/148)x100%=32.43%

%N=(2.14/148)x100%=18.92%

%O=(6.16/148)x100%=64.86%

Answer:

sullage or grey water

Explanation:

sources include: sinks, showers, baths, washing machines, dishwashers and no not toliets.

Answer:

The frequency of coral diseases has increased significantly over the last 10 years, causing widespread mortality among reef-building corals. Many scientists believe the increase is related to deteriorating water quality associated with human-made pollutants and increased sea surface temperatures.

Explanation:

Answer:

When one object is rubbed against another, static electricity can be created. This is because the rubbing creates a negative charge that is carried by electrons. The electrons can build up to produce static electricity. For example, when you shuffle your feet across a carpet, you are creating many surface contacts between your feet and the carpet, allowing electrons to transfer to you, thereby building up a static charge on your skin. When you touch another person or an object, you can suddenly discharge the static as an electrical shock.

Similarly, when you rub a balloon on your head it causes opposite static charges to build up both on your hair and the balloon. Consequently, when you pull the balloon slowly away from your head, you can see these two opposite static charges attracting one another and making your hair stand up.

Materials

• Balloon

• An object made out of wool (such as a sweater, scarf, blanket or ball of yarn)

• Stopwatch

• A wall

• A partner (optional)

Preparation

• Blow up the balloon and tie off the end.

• Have your partner prepare to use the stopwatch.

Procedure

• Hold the balloon in a way that your hand covers as little of its surface area as possible, such as by using only your thumb and pointer finger or by gripping the balloon by its neck where it is tied off.

• Rub the balloon on the woolly object once, in one direction.

• Hold the balloon up on the wall with the side that was rubbed against the wool facing the wall, then release it. Does the balloon stay stuck on the wall? If the balloon stays stuck, have your partner immediately start the stopwatch to time how long the balloon remains bound to the wall. If the balloon does not stick, move to the next step.

• Touch the balloon to a metal object. Why do you think this is important to do?

• Repeat the above process but each time increase the number of times you rub the balloon on the woolly object. Rub the balloon in the same direction each time. (Do not rub the balloon back and forth.)

Observations and results

In general, did the balloon stick to the wall for a longer amount of time as you increased the number of times you rubbed the balloon on the woolly object?

Wool is a conductive material, which means it readily gives away its electrons. Consequently, when you rub a balloon on wool, this causes the electrons to move from the wool to the balloon's surface. The rubbed part of the balloon now has a negative charge. Objects made of rubber, such as the balloon, are electrical insulators, meaning that they resist electric charges flowing through them. This is why only part of the balloon may have a negative charge (where the wool rubbed it) and the rest may remain neutral.

When the balloon has been rubbed enough times to gain a sufficient negative charge, it will be attracted to the wall. Although the wall should normally have a neutral charge, the charges within it can rearrange so that a positively charged area attracts the negatively charged balloon. Because the wall is also an electrical insulator, the charge is not immediately discharged. However, because metal is an electrical conductor, when you rub the balloon against metal the extra electrons in the balloon quickly leave the balloon and move into the metal so the balloon is no longer attracted and does not adhere.

HOPE IT HELPS

PLEASE MARK ME BRAINLIEST

The test variables and the control variable and the outcome variables are Balloons, the electrical insulators, and resist electric charges flowing through them.

When we brush a balloon against our head, the hair and balloon both develop opposing static charges. We can see how these two opposing static charges attract one another and make our hair stand up as the balloon is gently dragged away from our head.

When a balloon is rubbed to the wall, it will be attracted to the wall. Because the wall is an electrical insulator, the charge is not quickly discharged.

The excess electrons in the balloon immediately leave the balloon and go into the metal when it is rubbed against metal. As a result, the balloon is no longer attracted to the metal and does not adhere.

Therefore, Balloons, which act as electrical insulators and prevent the flow of electric charges, serve as the test, control, and outcome variables.

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Answer: A

Explanation:

Answer:

C

Explanation:

because the combustion reaction produce vapor water and CO2

Answer:

5 g of H₂ are contained in the vessel

Explanation:

A typical excersise of conversion:

We determine molar mass of H₂.

This is a dyatomic molecule, as molar mass of H is 1g/mol, molar mass of H₂ will be 2g/mol.

Let's make the conversion of units:

2.5 mol . 2g/mol = 5 g

5 g of H₂ are contained in the vessel

Mass of hydrogen in the vessel = 5 grams

The number of moles is defined as the given mass over molar mass.

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Given:

Number of moles= 2.5 mol

To find:

Mass of hydrogen=?

Molar mass of hydrogen =2g/mol

On substituting the values in the given formula we will get:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\\\\text{Given mass}=\text{Number of moles}*\text{Molar mass}\\\\\text{Given mass}=2.5 mol* 2g/mol\\\\\text{Given mass}=5 g[/tex]

Thus, mass of hydrogen in the vessel is 5 grams.

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Answer:

Atoms

Explanation:

Answer:

33.6 i think

Explanation:

Answer:

u didnt give a worksheet

Explanation:

Answer:

1eV

Explanation:

Answer:

1ev

Explanation:

Reaction of galvanic cell

Given

Redox reaction between Zn and Mg

Required

Half reaction

Potential of the cell

Solution

Zn²⁺ + 2e -> Zn E ° = -0.76 V

Mg²⁺ + 2e -> Mg E ° = - 2.37 V

The higher E ° of the cell will act as the positive pole/cathode

Electrode Zn as a cathode (reduction) and Mg as an anode (oxidation) (E ° of Zn cells is greater than Mg)

Half reaction

Cathode: Zn²⁺ + 2e ⇒ Zn E ° = -0.76 V

Anode: Mg ⇒ Mg²⁺ + 2e E ° = +2.37 V

------------------------------------------------- ----------- +

Cell reaction: Zn²⁺ + Mg ---> Zn + Mg²⁺ E ° cell = +1.61 V

The reaction occurs spontaneously in the absence of an electric current, thus including galvanic cells/voltaic cells