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3 years ago

Choose the number of significant figures 60.0

Chemistry

2 answers:

GenaCL600 [577]3 years ago

8 0

Answer:

3 significant figures

Explanation:

The rules for deciding which digits in a measurement are significant are as follows:

1) All nonzero digits are significant.

2) Sandwiched (or embedded) zeros, those between significant digits, are significant.

3) Leading zeros, which are zeros at the beginning of a decimal number less than 1, are not significant.

4) Trailing zeros, which are zeros at the end of a number, are significant only if the number has a decimal point.

By rule 4 since 60.0 has a decimal point all 3 figures are significant

oksian1 [2.3K]3 years ago

4 0

In this case there are 3 significant figures.

Any zeroes after another number, (i.e. 9300.000), you consider these numbers.

If there is something like 0.0060, this would be two significant digits.

Zeroes before a number are not significant, anything after are.

Hope this helps any confusion!

Please mark me brainliest!

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The titration of 25.0 mL of an iron(II) solution required 18.0 mL of a 0.145 M solution of dichromate to reach the equivalence p

Svetlanka [38]

Answer:

0.64 M

Explanation:

Given:

Volume of iron(II) solution (V₁) = 25.0 mL = 0.025 L

Molarity of iron(II) solution (M₁) = ?

Number of moles of iron(II) solution (n₁) = ?

Volume of dichromate solution (V₂) = 18.0 mL = 0.018 L

Molarity of dichromate solution (M₂) = 0.145 M

Number of moles of dichromate solution (n₂) = ?

Molarity is equal to the ratio of moles and volume.

So, molarity of dichromate solution is given as:

$M_2=\frac{n_2}{V_2}\\\\n_2=M_2\times V_2=0.145\times 0.018 = 2.61\times 10^{-3}\ mol$

Now, let us write the complete balanced reaction for the given situation.

So, the complete balanced equation is given below.

$6Fe^{2+}(aq)+Cr_2O_7^{2-}(aq)+14H^+(aq)\to 6Fe^{3+}(aq)+2Cr^{3+}(aq)+7H_2O$

From the equation, it is clear that, 1 mole of dichromate is required for 6 moles of iron(II) solution.

So, using unitary method, we find the number of moles of iron(II) solution.

1 mole of dichromate = 6 moles of iron(II)

∴ n₂ moles of dichromate = 6n₂ moles of iron(II)

= $6\times 2.61\times 10^{-3}=0.016\ mol\ Fe^{2+}$

So, 0.016 moles of iron(II) is needed. Therefore, $n_1=0.016\ mol$

Now, molarity of iron(II) solution is given as:

Molarity = Moles ÷ Volume

$M_1=\frac{n_1}{V_1}\\\\M_1=\frac{0.016\ mol}{0.025\ L}=0.64\ M$

Therefore, the molarity of the iron(II) solution is 0.64 M.

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3 years ago

The combination of sulfur dioxide, nitrogen oxide, and atmospheric moisture creates

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This combination creates acid rain

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3 years ago

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If you have 30.O g of hydrogen gas burned in excess oxygen how many moles of water can you make

vladimir1956 [14]

Answer:

15 moles

Explanation:

Data given:

mass of hydrogen (H₂) = 30.0 g

amount of oxygen (O₂) = excess

moles of water = ?

Solution:

First we look to the reaction in which hydrogen react with oxygen and make (H₂O)

Reaction:

2H₂ + O₂ -----------> 2H₂O

Now look at the reaction for mole ratio

2H₂ + O₂ -----------> 2H₂O

2 mole 2 mole

So it is 2:2 mole ratio of hydrogen to water

As we Know

molar mass of H₂ = 2(1) = 2 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now convert moles to gram

2H₂ + O₂ -----------> 2H₂O

2 mole (2 g/mol) 2 mole (18 g/mol)

4 g 36 g

So,

we come to know that 4 g of hydrogen gives 36 g of water then how many grams of water will be produce by 30 grams of hydrogen.

Apply unity formula

4 g of H₂ ≅ 36 g of H₂O

30 g of H₂ ≅ X of H₂O

Do cross multiplication

X of H₂O = 30 g x 36 g / 4 g

X of H₂O = 270 g

Now convert grams of H₂O into moles

No. of moles = mass in grams/molar mass

Put values in above formula

No. of moles = 270 g / 18 (g/mol)

No. of moles = 15 mol

so 30 gram of hydrogen produce 15 mol of water.

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3 years ago

What is the total anion concentration (in mEq/L ) of a solution that contains 6.0 mEq/L Na + , 11.0 mEq/L Ca 2+ , and 1.0 mEq/L

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Which is not a type of mass movement

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Answer:

options?

Explanation:

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