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steposvetlana [31]
4 years ago
8

What reagent is used to test for starch​

Chemistry
1 answer:
Elodia [21]4 years ago
8 0

Answer:

iodine

Explanation:

In the presence of starch, iodine turns a blue/black colour. It is possible to distinguish starch from glucose (and other carbohydrates) using this iodine solution test. For example, if iodine is added to a peeled potato then it will turn black. Benedict's reagent can be used to test for glucose.

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B) causes new mutations to occur in the offspring.

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What is a substance?
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I think the answer is C. An element or compound that cannot be physically separated. Sorry if im wrong.

7 0
4 years ago
NaOH + X mc005-1.jpg NaCH3COO + H2O What is X in this reaction?
Allushta [10]
"X" in the reaction above is acetic acid with a chemical formula CH3COOH. The chemical reaction would be NaOH + CH3COOH = NaCH3COO + H2O. This is a neutralization reaction in which it produces a salt and water. The salt produced is called sodium acetate.
7 0
3 years ago
A gas system has volume, moles and temperature of 9040 mL, 0.447 moles and -35.50 oC, respectively. What is the pressure in atm?
babymother [125]

Answer : The pressure of the gas is, 0.964 atm

Solution : Given,

Volume of gas = 9040 ml = 9.040 L        (1 L = 1000 ml)

Moles of gas = 0.447 moles

Temperature of gas = -35.50^oC=237.5K       (0^oC=273K)

Using ideal gas equation,

PV=nRT

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = gas constant = 0.0821Latm/moleK

Now put all the given values in this formula, we get the pressure of the gas.

P(9.040L)=(0.447moles)\times (0.0821Latm/moleK)\times (237.5K)

By rearranging the terms, we get

P=0.964atm

Therefore, the pressure of the gas is, 0.964 atm

7 0
3 years ago
At high temperatures, carbon reacts with O2 to produce CO as follows: C(s) O2(g) 2CO(g). When 0.350 mol of O2 and excess carbon
777dan777 [17]

Answer:

Value of K_{c} is 0.090.

Explanation:

Initial molarity of O_{2} = \frac{0.350}{5.00}M = 0.0700 M

Construct an ICE table corresponding to the combustion reaction of carbon to determine K_{c}

                       C(s)+O_{2}(g)\rightarrow 2CO(g)

              I (M):    -      0.0700        0

             C (M):  -          -x             +2x

             E (M):   -    0.0700-x       2x

So, K_{c}=\frac{[CO]^{2}}{[O_{2}]}  , where [CO] and [O_{2}] represents equilibrium concentration of CO and O_{2} respectively.

Here, [CO]=2x=0.060

       ⇒x = 0.030

So, [O_{2}] = 0.0700-x = (0.0700-0.030) = 0.040

Hence,  K_{c}=\frac{(0.060)^{2}}{0.040}=0.090

4 0
4 years ago
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