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4 years ago

Which type of chemical reaction occurs in Ca+CI2 CaCI2

Chemistry

1 answer:

dimaraw [331]4 years ago

3 0

Ca + Cl2 = CaCl₂

A synthesis <span>reaction.</span>

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Calculate the theoretical yield for the bromination of both stilbenes

podryga [215]

Answer:

cinnamic acid - 150 mg

cis-stilbene - 100 μL

trans- stilbene - 100 mg

pyridinium tribromide - 200-385 mg

For this data:

moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols

Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g

cis-stilbene (100 ul = 0.1 ml)

moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols

Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g

trans-stilbene

moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols

Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g

Explanation:

4 0

2 years ago

What is the difference between accuracy and precision in chemistry ?

earnstyle [38]

Accuracy is more of a range

6 0

4 years ago

Suppose that on a dry, sunny day when the air temperature is near 37 ∘C,37 ∘C, a certain swimming pool would increase in tempera

ioda

Answer:

The fraction of water body necessary to keep the temperature constant is 0,0051.

Explanation:

Heat:

$Q=m*Ce*ΔT$

Q= heat (unknown)

m= mass (unknown)

Ce= especific heat (1 cal/g*°C)

ΔT= variation of temperature (2.75 °C)

Latent heat:

$ΔE=∝mΔHvap$

ΔE= latent heat

m= mass (unknown)

∝= mass fraction (unknown)

ΔHvap= enthalpy of vaporization (539.4 cal/g)

Since Q and E are equal, we can match both equations:

$m*Ce*ΔT=∝*m*ΔHvap$

Mass fraction is:

$∝=\frac{Ce*ΔT}{ΔHvap}$

$∝=\frac{(1 cal/g*°C)*2.75°C}{539.4 cal/g}$

∝=0,0051

7 0

3 years ago

What is the volume, in liters, of 6.8 mol of Kr gas at STP?

Virty [35]

Answer:

$\boxed {\boxed {\sf D. \ 152 \ L}}$

Explanation:

Any gas at standard temperature and pressure (STP) has a volume of 22.4 liters per mole or 22.4 L/mol. We can create a proportion with this value.

$\frac {22.4 \ L \ Kr}{1 \ mol \ Kr}$

Multiply both sides of the equation by 6.8 moles of krypton.

$6.8 \ mol \ Kr *\frac {22.4 \ L \ Kr}{1 \ mol \ Kr}$

The units of moles of krypton will cancel.

$6.8 *\frac {22.4 \ L \ Kr}{1 }$

The denominator of 1 can be ignored, so this becomes a simple multiplication problem.

$68 * 22.4 \ L \ Kr$

$152.32 \ L \ Kr$

If we round to the nearest whole number, the 3 in the tenths place tells us to leave the 2 in the ones place.

$152 \ L \ Kr$

6.8 moles of krypton gas at standard temperature and pressure is equal to <u>152 liters</u>.

6 0

3 years ago

what is the total ionic and net ionic equation for for the reaction between between mercury (II) nitrate and iron (III) chloride

MArishka [77]

Answer:

Hg+2 + 2Cl- -> HgCl2

HOPE THIS HELPS HAVE A GREAT DAY/NIGHT!!~

Explanation:

7 0

2 years ago