All categories

3 years ago

If a liquid has a density of 1200 g/l what is it’s density in charge/ml

Chemistry

1 answer:

Juliette [100K]3 years ago

5 0

Answer:- $1.2\frac{g}{mL}$

Solution:- The given density is $1200\frac{g}{L}$ and it asks to convert it to $\frac{g}{mL}$ .

Here, we could notice that only volume unit is changing ad we know that, 1 L = 1000 mL. Let's make the set up as:

$1200\frac{g}{L}(\frac{1L}{1000mL})$

= $1.2\frac{g}{mL}$

So, the density is $1.2\frac{g}{mL}$ .

You might be interested in

How many moles are in 29.5 grams of Ax?

JulsSmile [24]

The number of moles present in 29.5 grams of argon is 0.74 mole.

The atomic mass of argon is given as;

Ar = 39.95 g/mole

The number of moles present in 29.5 grams of argon is calculated as follows;

39.95 g ------------------------------- 1 mole

29.5 g ------------------------------ ?

$= \frac{29.5}{39.95} \\\\= 0.74 \ mole$

Thus, the number of moles present in 29.5 grams of argon is 0.74 mole.

<em>"Your question seems to be missing the correct symbol for the element" </em>

Argon = Ar

Learn more here:brainly.com/question/4628363

6 0

2 years ago

2C2H2 + 5O2 → 4CO2 + 2 H2O How many grams of CO2 are required to react with exactly 3.00 mol of O2? 106 g 132 g 165 g 76.8 g

cricket20 [7]

Hey there!:

Given the reaction:

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

5 moles O2 ------------- 4 moles CO2

3.00 moles O2 ---------- ( moles of CO2 ?? )

moles of CO2 = 3.00 * 4 / 5

moles of CO2 = 12 / 5

moles of CO2 = 2.4 moles

So, molar mass CO2 = 44.01 g/mol

Therefore:

1 mole CO2 -------------- 44.01 g

2.4 moles CO2 ---------- ( mass of CO2 )

mass of CO2 = 2.4 * 44.01 / 1

mass of CO2 = 106 g

Answer A

Hope that helps!

8 0

3 years ago

Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.7 mm long and 9.72 mm in diameter is stresse

amm1812

Answer:

The true stress required = 379 MPa

Explanation:

True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,

σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent

True strain is given by

Epsilon t =㏑ (l/l₀)

Substitute㏑(l/l₀) for ε(t)

σ(t) = K(㏑(l/l₀))ⁿ

Given values l₀ = 49.7mm, l =51.7mm , n =0.2 , σ(t) =379Mpa

379 x 10⁶ = K (㏑(51.7/49.7))^0.2

K = 379 x 10⁶/(㏑(51.7/49.7))^0.2

K = 723.48 MPa

Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.

σ(t) = K(㏑(l/l₀))ⁿ

l₀ = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa

σ(t) = 723.48 x 106 x (㏑(51.7/49.7))^0.2

σ(t) = 379 MPa

The true stress necessary to plastically elongate the specimen is 379 MPa.

6 0

3 years ago

Read 2 more answers

The nation on average uses around 10 Million tons of salt per year to de-ice roads. How many kilograms of salt is this? (please

Westkost [7]

45359237 kilograms :))))))))))

3 0

3 years ago

Methanol can be produced by the following reaction: CO(g) 2 H2(g) CH3OH(g). How is the rate of disappearance of hydrogen gas rel

belka [17]

Answer:

$r_{H_2}=-2r_{CH_3OH}$

Explanation:

Hello!

In this case, for the reaction:

$CO(g)+ 2 H_2(g) \rightarrow CH_3OH(g)$

In such a way, via the rate proportions, that is written considering the stoichiometric coefficients, we obtain:

$\frac{1}{-1} r_{CO}=\frac{1}{-2} r_{H_2}=\frac{1}{1} r_{CH_3OH}$

Whereas the reactants, CO and H2 have negative stoichiometric coefficients; therefore the rate of disappearance of hydrogen gas is related to the rate of appearance of methanol as shown below:

$\frac{1}{-2} r_{H_2}=\frac{1}{1} r_{CH_3OH}\\\\r_{H_2}=\frac{-2}{1} r_{CH_3OH}\\\\r_{H_2}=-2r_{CH_3OH}$

Which means that the rate of disappearance of hydrogen gas is negative and the rate of appearance of methanol is positive.

Regards!

7 0

3 years ago