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3 years ago

Do you think it matters if you

Chemistry

1 answer:

vovikov84 [41]3 years ago

5 0

Answer:

yes I do think they mean the same thing

Explanation:

150 grams and 150 grams is the same thing and adding 0 to the end of a decimal does not change its value, you could even put 150.0000000 grams and it would still be equivalent to the other numbers

You might be interested in

Which of the statements below best describes a metallic bond?

zhenek [66]

I believe a. is the answer.

5 0

4 years ago

A type of marble has a density of 2.5 grams/marble. A box filled with marbles has a total mass of 300 grams determine the number

Maurinko [17]

Answer:

100

Explanation:

Step 1: Subtract 50 by 300 to find the total mass of the marbles

300 - 50 = 250

Not hard, right?

Step 2: Divide 250 by 2.5

250 ÷ 2.5 = 100

And... 100 is your answer :)

Hope this helps :)

-jp524

4 0

3 years ago

What is the final pressure, in mmHg, for a gas in an aerosol can at an initial pressure of 1.40 atm at 12°C which is heated to 3

liubo4ka [24]

Answer:

b. 1150 mmHg

General Formulas and Concepts:

Chemistry - Gas Laws

Gay Lussac Law - $\frac{P_1}{T_1} =\frac{P_2}{T_2}$

P₁ is Pressure 1
T₁ is Temperature 1 in Kelvin
P₂ is Pressure 2
T₂ is Temperature 2 in Kelvin

Explanation:

Step 1: Define

P₁ = 1.40 atm

T₁ = 12°C

P₂ = unknown

T₂ = 35°C

Step 2: Identify Conversions

1 atm = 760 mmHg

K = °C + 273.15

Step 3: Convert

P₁ = 1.40 atm = 1064 mmHg

T₁ = 12°C = 285.15 K

T₂ = 35°C = 308.15 K

Step 4: Find P₂

Substitute: $\frac{1064 \ mmHg}{285.15 \ K} =\frac{P_2}{308.15 \ K}$
Cross-multiply: $(1064 \ mmHg)(308.15 \ K)=P_2(285.15 \ K)$
Multiply: $327872 \ mmHg \cdot K=P_2(285.15 \ K)$
Isolate P₂: $\frac{327872 \ mmHg \cdot K}{285.15 \ K}=P_2$
Divide: $1149.82 \ mmHg=P_2$
Rewrite: $P_2=1149.82 \ mmHg$

Step 5: Check

We are given 3 sig figs. Follow sig fig rules and round.

1149.82 mmHg ≈ 1150 mmHg

4 0

3 years ago

2Al + 6HCl --> 2AlCl3 + 3H2 Aluminium reacts with hydrochloric acid. How many grams of aluminum are necessary to produce 11 L

dem82 [27]

Answer:

8.8g of Al are necessaries

Explanation:

Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.

To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:

Moles H2:

PV = nRT; PV/RT = n

Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP

Replacing:

1atm*11L/0.082atmL/molK*273.15K = n

n = 0.491 moles of H2 must be produced

Moles Al:

0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required

Mass Al -Molar mass: 26.98g/mol-:

0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries

7 0

3 years ago

WHAT IN THE WORLD IS THIS?

AveGali [126]

Answer:

Explanation:

Nothing but the D

5 0

3 years ago