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3 years ago

What is the difference between a cation and an anion?

Chemistry

1 answer:

zmey [24]3 years ago

3 0

An anion has a negative charge, while a cation has a positive charge.

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Which of the following choices is a sign that a glacier was present at one point in history?

Alja [10]

answer:

B. U-shaped valleys

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3 years ago

(04.01 LC)

Marysya12 [62]

Answer:

Substance at the beginning of a reaction- reactant

Substance at the end of a reaction- product

Number placed before a compound in a chemical equation- stoichiometric coefficient

Explanation:

In a reaction equation, the species written on the left hand side of the equation are called the reactants.

The reactants combine to form the species on the right hand side of the reaction equation called products.

The stoichiometric coefficient is a number written before the formula of a compound in the reaction equation.

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3 years ago

For a tablet containing 500. mg of vitamin C, calculate how many ml of 0.095 M NaOH is required to reach the equivalence point.

Gala2k [10]

Answer:

mL of NaOH required =29.9mL

Explanation:

Let us calculate the moles of vitamin C in the tablet:

The molar mass of Vitamin C is 176.14 g/mole

$moles=\frac{mass}{molarmass}=\frac{500mg}{176.14}=\frac{0.5}{176.14}=0.0028$

Thus we need same number of moles of NaOH to reach the equivalence point.

For NaOH solution:

$moles=MolarityXvolume=0.095Xvolume$

$0.00283=0.095Xvolume$

$volume=0.0299L=29.9mL$

7 0

3 years ago

What are some geographic features that could be found in the hydrosphere?

nadezda [96]

Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.

hydro = water

6 0

3 years ago

An insulated container contains 0.3 kg of water at 20 degrees C. An alloy with a mass of 0.090 kg and an initial temperature of

Lorico [155]

Answer:

The specific heat of the alloy is 2.324 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of water = 0.3 kg = 300 grams

Temperature of water = 20°C

Mass of alloy = 0.090 kg

Initial temperature of alloy = 55 °C

The final temperature = 25°C

The specific heat of water = 4.184 J/g°C

<u>Step 2:</u> Calculate the specific heat of alloy

Qlost = -Qwater

Qmetal = -Qwater

Q = m*c*ΔT

m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)

⇒ mass of alloy = 90 grams

⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED

⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C

⇒ mass of water = 300 grams

⇒ c(water) = the specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C

90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C

c(alloy) = 2.324 J/g°C

The specific heat of the alloy is 2.324 J/g°C

3 0

3 years ago