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Georgia [21]
3 years ago
6

How do you write 145,000,000 in scientific notation?

Chemistry
1 answer:
belka [17]3 years ago
5 0

Answer:

\boxed{1.45 \times 10^{8}}

Explanation:

A number in scientific notation has the form N × 10ⁿ

where N is a decimal number called the mantissa and n is an integer called the exponent.

We must have 1 ≤ N < 10

Step 1. Determine the mantissa

Move the decimal place to the left to create a new number between 1 and 10.

145 000 000 ⟶ 1.450 000 00; N = 1.45

Step 2. Determine the exponent.

The exponent is the number of times you moved the decimal to get the mantissa.

You moved the decimal eight places to the left, so the exponent n = 8.

Step 3. Write the number in scientific notation

The number in scientific notation is \boxed{1.45 \times 10^{8}}.

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Zinc phosphate is used as a dental cement. A 50.00-mg sample is broken down into its constituent elements and gives 16.58 mg oxy
cricket20 [7]

Answer:

Zn3P2O8

Explanation:

In this particular question, it is necessary to convert the respective masses to percentages. We convert to percentages by placing each mass over the total mass and multiplying by 100%. Since the total is 50mg, conversion to percentage can be done by multiplying the masses by 2 as 100/50 is 2

For Oxygen = 16.58 * 2 = 33.16%

For phosphorus = 8.02 * 2 = 16.04%

For zinc = 25.40 * 2 = 50.80%

We then proceed to divide these percentages by their respective atomic masses. The atomic mass of oxygen, phosphorus and zinc are 16, 31 and 65 respectively.

O = 33.16/16 = 2.0725

P = 16.04/31 = 0.5174

Zn = 50.80/65 = 0.7815

Now, we divide by the smallest value which is that of the phosphorus

O = 2.0725/0.5174 = 4

P = 0.5174/0.5174 = 1

Zn= 0.7815/0.5174 = 1.5

Now, we need to multiply through by 2. This yields: O = 8, P = 2 and Zn = 3

The empirical formula is thus: Zn3P2O8

7 0
3 years ago
Help me plz with these problems
loris [4]
1 it a 2 it b 3 it c 4 it be 7 it a
6 0
3 years ago
Read 2 more answers
A gas is collected at 20.0 °C and 725.0 mm Hg. When the temperature is
krek1111 [17]

Answer:

676mmHg

Explanation:

Using the formula;

P1/T1 = P2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

P1 = 725.0mmHg

P2 = ?

T1 = 20°C = 20 + 273 = 293K

T2 = 0°C = 0 + 273 = 273K

Using P1/T1 = P2/T2

725/293 = P2/273

Cross multiply

725 × 273 = 293 × P2

197925 = 293P2

P2 = 197925 ÷ 293

P2 = 676mmHg.

The resulting pressure is 676mmHg

3 0
3 years ago
5. A penny weighs about 2.5 g. How many moles of pennies would be required to equal the mass of the moon (7.3x10^24 kg)
schepotkina [342]

5. 1.16 x 10^{26} moles  moles of pennies would be required to equal the mass of the moon.

6. 12.86 moles of ethanol are in a 750 ml bottle of vodka.

Explanation:

5 .Data given:

mass of penny = 2.5 grams

atomic mass of penny = 62.93 grams/mole

moles present in mass of the moon given as = 7.3 x 10^{24} kg

number of moles = \frac{mass}{atomic mass of  1 mole}

number of moles = \frac{2.5}{62.93}

0.039 moles of penny is present in 2.5 grams

0.039 moles of penny in 2.5 grams of it

so, x moles in 7.3 X 10^{27}  grams

\frac{0.039}{2.5} =\frac{x}{7.3 X 10^{27} }

x = 1.16 x 10^{26} moles

so when the mass of the penny given is equal to the mass of moon, number of moles of penny present is 1.1 x 10^{26} .

6.

Given:

vodka = 40% ethanol

volume of vodka bottle = 750 ml

moles of ethanol =?

density of ethanol =0.79 g/ml

atomic mass of ethanol = 46.07 grams/mole

so, from the density of ethanol given we can calculate how much ethanol is present in the solution.

density = \frac{mass}{volume}

density x volume = mass

0.79 x 750 = 592.5 grams

number of moles = \frac{mass}{atomic mass of 1 mole}

number of moles of ethanol = \frac{592.5}{46.07}

                                               = 12.86 moles of ethanol

6 0
3 years ago
3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,
VARVARA [1.3K]

Answer:

Cp_{liquid}=2.54\frac{J}{g\°C}

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

Q_{Ag}=-Q_{liquid}

That in terms of the heat capacities, masses and temperature changes turns out:

m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}

Best regards.

6 0
4 years ago
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