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olga55 [171]
1 year ago
7

What is 10 times as many as 1 hundred is blank hundred or blank thousand ?

Chemistry
1 answer:
Rama09 [41]1 year ago
8 0

10 times as many as 1 hundred is 10 hundred or 1 thousand.

This is the question of simple multiplications related to a place value system.

The given equation in unit form can be rewritten as:

10 × 1 × 100 = (X × 100) or (Y × 1000)

Arranging in two parts, we get,

10 × 1 × 100 = X × 100           ........(1)

10 × 1 × 100 = Y × 1000         ........(2)

Now, calculating the value of X and Y,

From equation (1), we get,

10 × 1 × 100 = X×100

or, 1000 = X × 100

or, X = 10

Hence, 10 times as many as 1 hundred is 10 hundred.

Similarly, from equation (2). we get,

10 × 1 × 100 = Y × 1000

or, 1000 = Y × 1000

or, Y = 1

Hence, 10 times as many as 1 hundred is 1 thousand.

Therefore, 10 times as many as 1 hundred is 10 hundred or 1 thousand.

To learn more about hundreds, thousands, and place values, visit: brainly.com/question/17785584

#SPJ9

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Which salt is the most common found in ocean water?
Ivenika [448]

Answer:

Sodium chloride

Explanation:

There are several salts in seawater, but the most abundant is ordinary table salt or sodium chloride (NaCl). Sodium chloride, like other salts, dissolves in water into its ions, so this is really a question about which ions are present in the greatest concentration. Sodium chloride dissociates into Na+ and Cl- ions.

7 0
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If the density of mercury is 13.6 g/mL, what is the mass of 25.4 cm3 of<br> mercury?
Anettt [7]

Answer:

345.44 g or 0.34544 kg

Explanation:

Applying

D = m/V...................... Equation 1

Where D = Density of mercury, m = mass of mercury, V = Volume of mercury.

make m the subject of the equation

m = D×V................. Equation 2

From the question,

Given: D = 13.6 g/mL = 13.6 g/cm³, V = 25.4 cm³

Substitute these values into equation 2

m = 13.6×25.4

m = 345.44 g

m = 0.34544 kg

Hence the mass of mercury is 345.44 g or 0.34544 kg

4 0
3 years ago
How many grams of oxygen are required to react completely with 3.6L of hydrogen at
Paladinen [302]
the number of moles of oxygen required are 0.08 mol. The volume of oxygen that is required to react can be calculated by the formula shown below. Substitute the values in equation (II). Hence, the volume of oxygen required to react with 3.6 L hydrogen is 1.8L . I hope this helps if not I’m sorry
7 0
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