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3 years ago

Value: 4

Chemistry

1 answer:

Komok [63]3 years ago

5 0

Answer:

x = 4.73

Explanation:

Given

y = 0.14x - 0.0398

Required

Find x when y = 0.622

To do this, we first substitute 0.622 for x.

y = 0.14x - 0.0398 gives

0.622 = 0.14x - 0.0398

Solve for 0.14x

0.14x = 0.622 + 0.0398

0.14x = 0.6618

Solve for x

x = 0.6618/0.14

x = 4.7271428571

x = 4.73 approximated

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What is the limiting reactant in the following equation? How much Fe2O3 will be produced if 2.1 g of Fe reacts with 2.1 g of O2?

Nuetrik [128]

Answer:

Fe is limiting reactant and 3.00g of Fe2O3 will be produced

Explanation:

To solve this question we must convert the mass of each reactant to moles and, using the reaction we can find limiting reactant. With moles of limiting reactant we can find moles of Fe2O3 and its mass as follows:

Moles Fe -Molar mass: 55.845g/mol-

2.1g * (1mol / 55.845g) = 0.0376 moles

Moles O2 -Molar mass: 32g/mol-

2.1g * (1mol / 32g) = 0.0656 moles

For a complete reaction of 0.0656 moles of O2 are needed:

0.0656moles O2 * (4mol Fe / 3 mol O2) = 0.0875 moles Fe

As there are just 0.0376 moles,

Fe is limiting reactant

The mass of Fe2O3 is:

Moles:

0.0376 moles Fe* (2mol Fe2O3 / 4mol Fe) = 0.0188 moles Fe2O3

Mass:

0.0188 moles Fe2O3 * (159.69g / mol) =

3.00g of Fe2O3 will be produced

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3 years ago

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mel-nik [20]

Answer:

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Explanation:

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3 years ago

Give the name and formula of the compound formed from the following elements: a) Sodium and nitrogen b) Oxygen and strontium c)

elena-s [515]

Answer: a) Sodium and nitrogen : $NaN_3$ : sodium nitride.

b) Oxygen and strontium : $SrO$ : strontium oxide.

c) Aluminum and chlorine : $AlCl_3$ : aluminium chloride.

d) Cesium and bromine : $CsBr$ : cesium bromide.

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

a) Sodium and nitrogen : Here sodium is having an oxidation state of +1 called as $Na^{+}$ cation and nitrogen forms an anion $N^{3-}$ with oxidation state of -3. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $NaN_3$ named as sodium nitride.

b) Oxygen and strontium : Here strontium is having an oxidation state of +2 called as $Sr^{+}$ cation and oxygen forms an anion $O^{2-}$ with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $SrO$ named as strontium oxide.

c) Aluminum and chlorine : Here aluminium is having an oxidation state of +3 called as $Al^{3+}$ cation and chlorine forms an anion $Cl^{-}$ with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $AlCl_3$ named as aluminium chloride.

d) Cesium and bromine : Here cesium is having an oxidation state of +1 called as $Cs^{+}$ cation and bromine forms an anion $Br^{-}$ with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $CsBr$ named as cesium bromide.